The textbook for the Data Science course series is freely available online.

Learning Objectives

Course Overview

Section 1: Introduction to Data Visualization and Distributions

You will get started with data visualization and distributions in R.

Section 2: Introduction to ggplot2

You will learn how to use ggplot2 to create plots.

Section 3: Summarizing with dplyr

You will learn how to summarize data using dplyr.

Section 4: Gapminder

You will see examples of ggplot2 and dplyr in action with the Gapminder dataset.

Section 5: Data Visualization Principles

You will learn general principles to guide you in developing effective data visualizations.

Section 1 Overview

Section 1 introduces you to Data Visualization and Distributions.

After completing Section 1, you will:

Introduction to Data Visualization

The textbook for this section is available here

Key points

Code

if(!require(dslabs)) install.packages("dslabs")
## Loading required package: dslabs
library(dslabs)
data(murders)
head(murders)
##        state abb region population total
## 1    Alabama  AL  South    4779736   135
## 2     Alaska  AK   West     710231    19
## 3    Arizona  AZ   West    6392017   232
## 4   Arkansas  AR  South    2915918    93
## 5 California  CA   West   37253956  1257
## 6   Colorado  CO   West    5029196    65

Introduction to Distributions

The textbook for this section is available here

Key points

Data Types

The textbook for this section is available here

Key points

Assessment - Data Types

  1. The type of data we are working with will often influence the data visualization technique we use.

We will be working with two types of variables: categorical and numeric. Each can be divided into two other groups: categorical can be ordinal or not, whereas numerical variables can be discrete or continuous.

We will review data types using some of the examples provided in the dslabs package. For example, the heights dataset.

library(dslabs)
data(heights)
data(heights)
names(heights)
## [1] "sex"    "height"
  1. We saw that sex is the first variable. We know what values are represented by this variable and can confirm this by looking at the first few entires:
head(heights)
##      sex height
## 1   Male     75
## 2   Male     70
## 3   Male     68
## 4   Male     74
## 5   Male     61
## 6 Female     65

What data type is the sex variable?

  1. Keep in mind that discrete numeric data can be considered ordinal.

Although this is technically true, we usually reserve the term ordinal data for variables belonging to a small number of different groups, with each group having many members.

The height variable could be ordinal if, for example, we report a small number of values such as short, medium, and tall. Let’s explore how many unique values are used by the heights variable. For this we can use the unique function:

x <- c(3, 3, 3, 3, 4, 4, 2)
unique(x)
x <- heights$height
length(unique(x))
## [1] 139
  1. One of the useful outputs of data visualization is that we can learn about the distribution of variables.

For categorical data we can construct this distribution by simply computing the frequency of each unique value. This can be done with the function table. Here is an example:

x <- c(3, 3, 3, 3, 4, 4, 2)
table(x)
x <- heights$height
tab <- table(x)
  1. To see why treating the reported heights as an ordinal value is not useful in practice we note how many values are reported only once.

In the previous exercise we computed the variable tab which reports the number of times each unique value appears. For values reported only once tab will be 1. Use logicals and the function sum to count the number of times this happens.

tab <- table(heights$height)
sum(tab==1)
## [1] 63
  1. Since there are a finite number of reported heights and technically the height can be considered ordinal, which of the following is true:

Describe Heights to ET

The textbook for this section is available:

Key points

Code

# load the dataset
library(dslabs)
data(heights)
# make a table of category proportions
prop.table(table(heights$sex))
## 
##    Female      Male 
## 0.2266667 0.7733333

Smooth Density Plots

The textbook for this section is available here

Key points

A further note on histograms: note that the choice of binwidth has a determinative effect on shape. There is no “true” choice for binwidth, and you can sometimes gain insights into the data by experimenting with binwidths.

Assessment - Distributions

  1. You may have noticed that numerical data is often summarized with the average value.

For example, the quality of a high school is sometimes summarized with one number: the average score on a standardized test. Occasionally, a second number is reported: the standard deviation. So, for example, you might read a report stating that scores were 680 plus or minus 50 (the standard deviation). The report has summarized an entire vector of scores with with just two numbers. Is this appropriate? Is there any important piece of information that we are missing by only looking at this summary rather than the entire list? We are going to learn when these 2 numbers are enough and when we need more elaborate summaries and plots to describe the data.

Our first data visualization building block is learning to summarize lists of factors or numeric vectors. The most basic statistical summary of a list of objects or numbers is its distribution. Once a vector has been summarized as distribution, there are several data visualization techniques to effectively relay this information. In later assessments we will practice to write code for data visualization. Here we start with some multiple choice questions to test your understanding of distributions and related basic plots.

In the murders dataset, the region is a categorical variable and on the right you can see its distribution. To the closest 5%, what proportion of the states are in the North Central region?

Region vs. Proportion

  1. In the murders dataset, the region is a categorical variable and to the right is its distribution.

Which of the following is true:

  1. The plot shows the eCDF for male heights.

Based on the plot, what percentage of males are shorter than 75 inches?

eCDF for male heights

  1. To the closest inch, what height m has the property that 1/2 of the male students are taller than m and 1/2 are shorter?
  1. Here is an eCDF of the murder rates across states.

eCDF of the murder rates across states

Knowing that there are 51 states (counting DC) and based on this plot, how many states have murder rates larger than 10 per 100,000 people?

  1. Based on the eCDF above, which of the following statements are true.
  1. Here is a histogram of male heights in our heights dataset.

Based on this plot, how many males are between 62.5 and 65.5?

Histogram of male heights

  1. About what percentage are shorter than 60 inches?
  1. Based on this density plot, about what proportion of US states have populations larger than 10 million?

Density plot population

  1. Below are three density plots. Is it possible that they are from the same dataset?

Three density plots

Which of the following statements is true?

Normal Distribution

The textbook for this section is available here

Key points

Equation for the normal distribution

The normal distribution is mathematically defined by the following formula for any mean \(\mu\) and standard deviation \(\sigma\):

\(Pr(a < x < b) = \int_{a}^{b} \frac{1}{\sqrt2\pi\sigma} e^-\frac{1}{2}(\frac{x - \mu}{\sigma})^2 dx\)

Code

if(!require(tidyverse)) install.packages("tidyverse")
## Loading required package: tidyverse
## ── Attaching packages ───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────── tidyverse 1.3.0 ──
## ✓ ggplot2 3.3.2     ✓ purrr   0.3.4
## ✓ tibble  3.0.3     ✓ dplyr   1.0.0
## ✓ tidyr   1.1.0     ✓ stringr 1.4.0
## ✓ readr   1.3.1     ✓ forcats 0.5.0
## ── Conflicts ──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()
# define x as vector of male heights
library(tidyverse)
index <- heights$sex=="Male"
x <- heights$height[index]

# calculate the mean and standard deviation manually
average <- sum(x)/length(x)
SD <- sqrt(sum((x - average)^2)/length(x))

# built-in mean and sd functions - note that the audio and printed values disagree
average <- mean(x)
SD <- sd(x)
c(average = average, SD = SD)
##   average        SD 
## 69.314755  3.611024
# calculate standard units
z <- scale(x)

# calculate proportion of values within 2 SD of mean
mean(abs(z) < 2)
## [1] 0.9495074

Note about the sd function: The built-in R function sd calculates the standard deviation, but it divides by length(x)-1 instead of length(x). When the length of the list is large, this difference is negligible and you can use the built-in sd function. Otherwise, you should compute \(\sigma\) by hand. For this course series, assume that you should use the sd function unless you are told not to do so.

Assessment - Normal Distribution

  1. Histograms and density plots provide excellent summaries of a distribution.

But can we summarize even further? We often see the average and standard deviation used as summary statistics: a two number summary! To understand what these summaries are and why they are so widely used, we need to understand the normal distribution.

The normal distribution, also known as the bell curve and as the Gaussian distribution, is one of the most famous mathematical concepts in history. A reason for this is that approximately normal distributions occur in many situations. Examples include gambling winnings, heights, weights, blood pressure, standardized test scores, and experimental measurement errors. Often data visualization is needed to confirm that our data follows a normal distribution.

Here we focus on how the normal distribution helps us summarize data and can be useful in practice.

One way the normal distribution is useful is that it can be used to approximate the distribution of a list of numbers without having access to the entire list. We will demonstrate this with the heights dataset.

Load the height data set and create a vector x with just the male heights:

library(dslabs)
data(heights)
x <- heights$height[heights$sex == "Male"]

What proportion of the data is between 69 and 72 inches (taller than 69 but shorter or equal to 72)? A proportion is between 0 and 1.

x <- heights$height[heights$sex == "Male"]
mean(x > 69 & x <= 72)
## [1] 0.3337438
  1. Suppose all you know about the height data from the previous exercise is the average and the standard deviation and that its distribution is approximated by the normal distribution.

We can compute the average and standard deviation like this:

library(dslabs)
data(heights)
x <- heights$height[heights$sex=="Male"]
avg <- mean(x)
stdev <- sd(x)

Suppose you only have avg and stdev below, but no access to x, can you approximate the proportion of the data that is between 69 and 72 inches?

Given a normal distribution with a mean mu and standard deviation sigma, you can calculate the proportion of observations less than or equal to a certain value with pnorm(value, mu, sigma). Notice that this is the CDF for the normal distribution. We will learn much more about pnorm later in the course series, but you can also learn more now with ?pnorm.

x <- heights$height[heights$sex=="Male"]
avg <- mean(x)
stdev <- sd(x)
pnorm(72, avg, stdev) - pnorm(69, avg, stdev)
## [1] 0.3061779
  1. Notice that the approximation calculated in the second question is very close to the exact calculation in the first question.

The normal distribution was a useful approximation for this case. However, the approximation is not always useful. An example is for the more extreme values, often called the “tails” of the distribution. Let’s look at an example. We can compute the proportion of heights between 79 and 81.

library(dslabs)  
data(heights)
x <- heights$height[heights$sex == "Male"]  
mean(x > 79 & x <= 81)  
x <- heights$height[heights$sex == "Male"]
avg <- mean(x)
stdev <- sd(x)
exact <- mean(x > 79 & x <= 81)
approx <- pnorm(81, avg, stdev) - pnorm(79, avg, stdev)
exact
## [1] 0.004926108
approx
## [1] 0.003051617
exact/approx
## [1] 1.614261
  1. Someone asks you what percent of seven footers are in the National Basketball Association (NBA). Can you provide an estimate? Let’s try using the normal approximation to answer this question.

First, we will estimate the proportion of adult men that are 7 feet tall or taller.

Assume that the distribution of adult men in the world as normally distributed with an average of 69 inches and a standard deviation of 3 inches.

# use pnorm to calculate the proportion over 7 feet (7*12 inches)
1 - pnorm(7*12, 69, 3)
## [1] 2.866516e-07
  1. Now we have an approximation for the proportion, call it p, of men that are 7 feet tall or taller.

We know that there are about 1 billion men between the ages of 18 and 40 in the world, the age range for the NBA.

Can we use the normal distribution to estimate how many of these 1 billion men are at least seven feet tall?

p <- 1 - pnorm(7*12, 69, 3)
round(p*10^9)
## [1] 287
  1. There are about 10 National Basketball Association (NBA) players that are 7 feet tall or higher.
p <- 1 - pnorm(7*12, 69, 3)
N <- round(p*10^9)
10/N
## [1] 0.03484321
  1. In the previous exerceise we estimated the proportion of seven footers in the NBA using this simple code:
p <- 1 - pnorm(7*12, 69, 3)  
N <- round(p * 10^9)  
10/N

Repeat the calculations performed in the previous question for Lebron James’ height: 6 feet 8 inches. There are about 150 players, instead of 10, that are at least that tall in the NBA.

## Change the solution to previous answer
p <- 1 - pnorm(7*12, 69, 3)
N <- round(p * 10^9)
10/N
## [1] 0.03484321
p <- 1 - pnorm(6*12+8, 69, 3)
N <- round(p * 10^9)
150/N
## [1] 0.001220842
  1. In answering the previous questions, we found that it is not at all rare for a seven footer to become an NBA player.

What would be a fair critique of our calculations?

Quantile-Quantile Plots

The textbook for this section is available here

Key points

Code

# define x and z
index <- heights$sex=="Male"
x <- heights$height[index]
z <- scale(x)

# proportion of data below 69.5
mean(x <= 69.5)
## [1] 0.5147783
# calculate observed and theoretical quantiles
p <- seq(0.05, 0.95, 0.05)
observed_quantiles <- quantile(x, p)
theoretical_quantiles <- qnorm(p, mean = mean(x), sd = sd(x))

# make QQ-plot
plot(theoretical_quantiles, observed_quantiles)
abline(0,1)

# make QQ-plot with scaled values
observed_quantiles <- quantile(z, p)
theoretical_quantiles <- qnorm(p) 
plot(theoretical_quantiles, observed_quantiles)
abline(0,1)

Percentiles

The textbook for this section is available here

Key points

Boxplots

The textbook for this section is available here

Key points

Assessment - Quantiles, percentiles, and boxplots

  1. When analyzing data it’s often important to know the number of measurements you have for each category.
male <- heights$height[heights$sex=="Male"]
female <- heights$height[heights$sex=="Female"]
length(male)
## [1] 812
length(female)
## [1] 238
  1. Suppose we can’t make a plot and want to compare the distributions side by side. If the number of data points is large, listing all the numbers is inpractical. A more practical approach is to look at the percentiles. We can obtain percentiles using the quantile function like this
library(dslabs)
data(heights)
quantile(heights$height, seq(.01, 0.99, 0.01))
male <- heights$height[heights$sex=="Male"]
female <- heights$height[heights$sex=="Female"]
female_percentiles <- quantile(female, seq(0.1, 0.9, 0.2))
male_percentiles <- quantile(male, seq(0.1, 0.9, 0.2))
df <- data.frame(female = (female_percentiles), male = (male_percentiles))
df
##       female     male
## 10% 61.00000 65.00000
## 30% 63.00000 68.00000
## 50% 64.98031 69.00000
## 70% 66.46417 71.00000
## 90% 69.00000 73.22751
  1. Study the boxplots summarizing the distributions of populations sizes by country.

Continent vs Population

Which continent has the country with the largest population size?

  1. Study the boxplots summarizing the distributions of populations sizes by country.

Which continent has median country with the largest population?

  1. Again, look at the boxplots summarizing the distributions of populations sizes by country.

To the nearest million, what is the median population size for Africa?

  1. Examine the following boxplots and report approximately what proportion of countries in Europe have populations below 14 million?
  1. Based on the boxplot, if we use a log transformation, which continent shown below has the largest interquartile range?

Distribution of Female Heights

The textbook for this section is available here

Key points

Assessment - Robust Summaries With Outliers

  1. For this chapter, we will use height data collected by Francis Galton for his genetics studies. Here we just use height of the children in the dataset:
library(HistData)
data(Galton)
x <- Galton$child
if(!require(HistData)) install.packages("HistData")
## Loading required package: HistData
## Warning: package 'HistData' was built under R version 4.0.2
library(HistData)
data(Galton)
x <- Galton$child
mean(x)
## [1] 68.08847
median(x)
## [1] 68.2
  1. Now for the same data compute the standard deviation and the median absolute deviation (MAD).
x <- Galton$child
sd(x)
## [1] 2.517941
mad(x)
## [1] 2.9652
  1. In the previous exercises we saw that the mean and median are very similar and so are the standard deviation and MAD. This is expected since the data is approximated by a normal distribution which has this property.

Now suppose that suppose Galton made a mistake when entering the first value, forgetting to use the decimal point. You can imitate this error by typing:

library(HistData)
data(Galton)
x <- Galton$child
x_with_error <- x
x_with_error[1] <- x_with_error[1]*10

The data now has an outlier that the normal approximation does not account for. Let’s see how this affects the average.

x <- Galton$child
x_with_error <- x
x_with_error[1] <- x_with_error[1]*10
gem <- mean(x)
gem_error <- mean(x_with_error)
gem_error - gem
## [1] 0.5983836
  1. In the previous exercise we saw how a simple mistake in 1 out of over 900 observations can result in the average of our data increasing more than half an inch, which is a large difference in practical terms.

Now let’s explore the effect this outlier has on the standard deviation.

x_with_error <- x
x_with_error[1] <- x_with_error[1]*10
sd(x_with_error)- sd(x)
## [1] 15.6746
  1. In the previous exercises we saw how one mistake can have a substantial effect on the average and the standard deviation.

Now we are going to see how the median and MAD are much more resistant to outliers. For this reason we say that they are robust summaries.

x_with_error <- x
x_with_error[1] <- x_with_error[1]*10
mediaan <- median(x)
mediaan_error <- median(x_with_error)
mediaan_error - mediaan
## [1] 0
  1. We saw that the median barely changes. Now let’s see how the MAD is affected.

We saw that the median barely changes. Now let’s see how the MAD is affected.

x_with_error <- x
x_with_error[1] <- x_with_error[1]*10
mad_normal <- mad(x)
mad_error <- mad(x_with_error)
mad_error - mad_normal
## [1] 0
  1. How could you use exploratory data analysis to detect that an error was made?
  1. We have seen how the average can be affected by outliers.

But how large can this effect get? This of course depends on the size of the outlier and the size of the dataset.

To see how outliers can affect the average of a dataset, let’s write a simple function that takes the size of the outlier as input and returns the average.

x <- Galton$child
error_avg <- function(k){
x[1] = k
mean(x)
}
error_avg(10000)
## [1] 78.79784
error_avg(-10000)
## [1] 57.24612

Section 2 Overview

In Section 2, you will learn how to create data visualizations in R using ggplot2.

After completing Section 2, you will:

Note that it can be hard to memorize all of the functions and arguments used by ggplot2, so we recommend that you have a cheat sheet handy to help you remember the necessary commands.

ggplot

The textbook for this section is available here

Key points

Graph Components

The textbook for this section is available here

Key points

Creating a New Plot

The textbook for this section is available here

Key points

Code

ggplot(data = murders)

murders %>% ggplot()
p <- ggplot(data = murders)
class(p)
## [1] "gg"     "ggplot"
print(p)    # this is equivalent to simply typing p

The functions above render a plot, in this case a blank slate since no geometry has been defined. The only style choice we see is a grey background.

Layers

The textbook for this section is available:

Key points

Code: Adding layers to a plot

murders %>% ggplot() +
    geom_point(aes(x = population/10^6, y = total))

# add points layer to predefined ggplot object
p <- ggplot(data = murders)
p + geom_point(aes(population/10^6, total))

# add text layer to scatterplot
p + geom_point(aes(population/10^6, total)) +
    geom_text(aes(population/10^6, total, label = abb))

Code: Example of aes behavior

# no error from this call
p_test <- p + geom_text(aes(population/10^6, total, label = abb))
# error - "abb" is not a globally defined variable and cannot be found outside of aes
p_test <- p + geom_text(aes(population/10^6, total), label = abb)

Tinkering

The textbook for this section is available here and here

Key points

Code

# change the size of the points
p + geom_point(aes(population/10^6, total), size = 3) +
    geom_text(aes(population/10^6, total, label = abb))

# move text labels slightly to the right
p + geom_point(aes(population/10^6, total), size = 3) +
    geom_text(aes(population/10^6, total, label = abb), nudge_x = 1)

# simplify code by adding global aesthetic
p <- murders %>% ggplot(aes(population/10^6, total, label = abb))
p + geom_point(size = 3) +
    geom_text(nudge_x = 1.5)

# local aesthetics override global aesthetics
p + geom_point(size = 3) +
    geom_text(aes(x = 10, y = 800, label = "Hello there!"))

Scales, Labels, and Colors

The textbook for this section is available:

Key points

Code: Log-scale the x- and y-axis

# define p
p <- murders %>% ggplot(aes(population/10^6, total, label = abb))

# log base 10 scale the x-axis and y-axis
p + geom_point(size = 3) +
    geom_text(nudge_x = 0.05) +
    scale_x_continuous(trans = "log10") +
    scale_y_continuous(trans = "log10")

# efficient log scaling of the axes
p + geom_point(size = 3) +
    geom_text(nudge_x = 0.075) +
    scale_x_log10() +
    scale_y_log10()

Code: Add labels and title

p + geom_point(size = 3) +
    geom_text(nudge_x = 0.075) +
    scale_x_log10() +
    scale_y_log10() +
    xlab("Population in millions (log scale)") +
    ylab("Total number of murders (log scale)") +
    ggtitle("US Gun Murders in 2010")

Code: Change color of the points

# redefine p to be everything except the points layer
p <- murders %>%
    ggplot(aes(population/10^6, total, label = abb)) +
    geom_text(nudge_x = 0.075) +
    scale_x_log10() +
    scale_y_log10() +
    xlab("Population in millions (log scale)") +
    ylab("Total number of murders (log scale)") +
    ggtitle("US Gun Murders in 2010")

# make all points blue
p + geom_point(size = 3, color = "blue")

# color points by region
p + geom_point(aes(col = region), size = 3)

Code: Add a line with average murder rate

# define average murder rate
r <- murders %>%
    summarize(rate = sum(total) / sum(population) * 10^6) %>%
    pull(rate)
    
# basic line with average murder rate for the country
p + geom_point(aes(col = region), size = 3) +
    geom_abline(intercept = log10(r))    # slope is default of 1

# change line to dashed and dark grey, line under points
p + 
    geom_abline(intercept = log10(r), lty = 2, color = "darkgrey") +
    geom_point(aes(col = region), size = 3)

Code: Change legend title

p <- p + scale_color_discrete(name = "Region")    # capitalize legend title

Add-on Packages

The textbook for this section is available here and here

Key points

Code: Adding themes

if(!require(ggthemes)) install.packages("ggthemes")
## Loading required package: ggthemes
## Warning: package 'ggthemes' was built under R version 4.0.2
# theme used for graphs in the textbook and course
ds_theme_set()

# themes from ggthemes
library(ggthemes)
p + theme_economist()    # style of the Economist magazine

p + theme_fivethirtyeight()    # style of the FiveThirtyEight website

Code: Putting it all together to assemble the plot

if(!require(ggrepel)) install.packages("ggrepel")
## Loading required package: ggrepel
## Warning: package 'ggrepel' was built under R version 4.0.2
# load libraries
library(ggrepel)

# define the intercept
r <- murders %>%
    summarize(rate = sum(total) / sum(population) * 10^6) %>%
    .$rate
    
# make the plot, combining all elements
murders %>%
    ggplot(aes(population/10^6, total, label = abb)) +
    geom_abline(intercept = log10(r), lty = 2, color = "darkgrey") +
    geom_point(aes(col = region), size = 3) +
    geom_text_repel() +
    scale_x_log10() +
    scale_y_log10() +
    xlab("Population in millions (log scale)") +
    ylab("Total number of murders (log scale)") +
    ggtitle("US Gun Murders in 2010") +
    scale_color_discrete(name = "Region") +
    theme_economist()

Other Examples

The textbook for this section is available:

Key points

Code: Histograms in ggplot2

# define p
p <- heights %>%
    filter(sex == "Male") %>%
    ggplot(aes(x = height))
    
# basic histograms
p + geom_histogram()
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

p + geom_histogram(binwidth = 1)

# histogram with blue fill, black outline, labels and title
p + geom_histogram(binwidth = 1, fill = "blue", col = "black") +
    xlab("Male heights in inches") +
    ggtitle("Histogram")

Code: Smooth density plots in ggplot2

p + geom_density()

p + geom_density(fill = "blue")

Code: Quantile-quantile plots in ggplot2

# basic QQ-plot
p <- heights %>% filter(sex == "Male") %>%
    ggplot(aes(sample = height))
p + geom_qq()

# QQ-plot against a normal distribution with same mean/sd as data
params <- heights %>%
    filter(sex == "Male") %>%
    summarize(mean = mean(height), sd = sd(height))
p + geom_qq(dparams = params) +
    geom_abline()

# QQ-plot of scaled data against the standard normal distribution
heights %>%
    ggplot(aes(sample = scale(height))) +
    geom_qq() +
    geom_abline()

Code: Grids of plots with the grid.extra package

if(!require(gridExtra)) install.packages("gridExtra")
## Loading required package: gridExtra
## 
## Attaching package: 'gridExtra'
## The following object is masked from 'package:dplyr':
## 
##     combine
# define plots p1, p2, p3
p <- heights %>% filter(sex == "Male") %>% ggplot(aes(x = height))
p1 <- p + geom_histogram(binwidth = 1, fill = "blue", col = "black")
p2 <- p + geom_histogram(binwidth = 2, fill = "blue", col = "black")
p3 <- p + geom_histogram(binwidth = 3, fill = "blue", col = "black")

# arrange plots next to each other in 1 row, 3 columns
library(gridExtra)
grid.arrange(p1, p2, p3, ncol = 3)

Assessment - ggplot2

  1. Start by loading the dplyr and ggplot2 libraries as well as the murders data.
library(dplyr)
library(ggplot2)
library(dslabs)
data(murders)

Note that you can load both dplyr and ggplot2, as well as other packages, by installing and loading the tidyverse package.

With ggplot2 plots can be saved as objects. For example we can associate a dataset with a plot object like this

p <- ggplot(data = murders)

Because data is the first argument we don’t need to spell it out. So we can write this instead:

p <- ggplot(murders)

or, if we load dplyr, we can use the pipe:

p <- murders %>% ggplot()

Remember the pipe sends the object on the left of %>% to be the first argument for the function the right of %>%.

Now let’s get an introduction to ggplot.

if(!require(dplyr)) install.packages("dplyr")

library(dplyr)
p <- ggplot(murders)
class(p)
## [1] "gg"     "ggplot"
  1. Remember that to print an object you can use the command print or simply type the object. For example, instead of
x <- 2
print(x)

you can simply type

x <-2
x

Print the object p defined in exercise one

p <- ggplot(murders)

and describe what you see.

  1. Now we are going to review the use of pipes by seeing how they can be used with ggplot.
# define ggplot object called p like in the previous exercise but using a pipe 
p <- heights %>% ggplot()
p # a blank slate plot

  1. Now we are going to add layers and the corresponding aesthetic mappings. For the murders data, we plotted total murders versus population sizes in the videos.

Explore the murders data frame to remind yourself of the names for the two variables (total murders and population size) we want to plot and select the correct answer.

  1. To create a scatter plot, we add a layer with the function geom_point.

The aesthetic mappings require us to define the x-axis and y-axis variables respectively. So the code looks like this:

murders %>% ggplot(aes(x = , y = )) +
  geom_point()

except we have to fill in the blanks to define the two variables x and y.

## Fill in the blanks
murders %>% ggplot(aes(x =population , y =total )) +
  geom_point()

  1. Note that if we don’t use argument names, we can obtain the same plot by making sure we enter the variable names in the desired order.
murders %>% ggplot(aes(population, total)) +
  geom_point()

  1. If instead of points we want to add text, we can use the geom_text() or geom_label() geometries.

However, note that the following code

murders %>% ggplot(aes(population, total)) +
  geom_label()

will give us the error message: Error: geom_label requires the following missing aesthetics: label

Why is this?

  1. You can also add labels to the points on a plot.
## edit the next line to add the label
murders %>% ggplot(aes(population, total, label = abb)) + geom_label()

  1. Now let’s change the color of the labels to blue. How can we do this?
  1. Now let’s go ahead and make the labels blue. We previously wrote this code to add labels to our plot:
murders %>% ggplot(aes(population, total, label= abb)) +
  geom_label()

Now we will edit this code.

murders %>% ggplot(aes(population, total,label= abb)) +
  geom_label(color="blue")

  1. Now suppose we want to use color to represent the different regions.

So the states from the West will be one color, states from the Northeast another, and so on.

In this case, which of the following is most appropriate:

  1. We previously used this code to make a plot using the state abbreviations as labels:
murders %>% ggplot(aes(population, total, label = abb)) +
  geom_label()

We are now going to add color to represent the region.

## edit this code
murders %>% ggplot(aes(population, total, label = abb, color=region)) +
  geom_label()

  1. Now we are going to change the axes to log scales to account for the fact that the population distribution is skewed.

Let’s start by defining an object p that holds the plot we have made up to now:

p <- murders %>% ggplot(aes(population, total, label = abb, color = region)) +
  geom_label()

To change the x-axis to a log scale we learned about the scale_x_log10() function. We can change the axis by adding this layer to the object p to change the scale and render the plot using the following code:

p + scale_x_log10()
p <- murders %>% ggplot(aes(population, total, label = abb, color = region)) + geom_label()
## add layers to p here
p + scale_x_log10() + scale_y_log10()

  1. In the previous exercises we created a plot using the following code:
library(dplyr)
library(ggplot2)
library(dslabs)
data(murders)
p<- murders %>% ggplot(aes(population, total, label = abb, color = region)) +
  geom_label()
p + scale_x_log10() + scale_y_log10()

We are now going to add a title to this plot. We will do this by adding yet another layer, this time with the function ggtitle.

p <- murders %>% ggplot(aes(population, total, label = abb, color = region)) + geom_label()
# add a layer to add title to the next line
p + scale_x_log10() + scale_y_log10() + ggtitle("Gun murder data")

  1. We are going to shift our focus from the murders dataset to explore the heights dataset.

We use the geom_histogram function to make a histogram of the heights in the heights data frame. When reading the documentation for this function we see that it requires just one mapping, the values to be used for the histogram.

What is the variable containing the heights in inches in the heights data frame?

  1. We are now going to make a histogram of the heights so we will load the heights dataset.

The following code has been pre-run for you to load the heights dataset:

library(dplyr)
library(ggplot2)
library(dslabs)
data(heights)
# define p here
p <- heights %>% ggplot(aes(height))
  1. Now we are ready to add a layer to actually make the histogram.
p <- heights %>% 
  ggplot(aes(height))
## add a layer to p
p + geom_histogram()
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

  1. Note that when we run the code from the previous exercise we get the following warning:
stat_bin() using bins = 30. Pick better value with binwidth.
p <- heights %>% 
  ggplot(aes(height))
## add the geom_histogram layer but with the requested argument
p + geom_histogram(binwidth = 1)

  1. Now instead of a histogram we are going to make a smooth density plot.

In this case, we will not make an object p. Instead we will render the plot using a single line of code. In the previous exercise, we could have created a histogram using one line of code like this:

heights %>% 
  ggplot(aes(height)) +
  geom_histogram()
## add the correct layer using +
heights %>% 
  ggplot(aes(height)) + geom_density()

  1. Now we are going to make density plots for males and females separately.

We can do this using the group argument within the aes mapping. Because each point will be assigned to a different density depending on a variable from the dataset, we need to map within aes.

## add the group argument then a layer with +
heights %>% 
  ggplot(aes(height, group = sex)) + geom_density()

  1. In the previous exercise we made the two density plots, one for each sex, using:
heights %>% 
  ggplot(aes(height, group = sex)) + 
  geom_density()

We can also assign groups through the color or fill argument. For example, if you type color = sex ggplot knows you want a different color for each sex. So two densities must be drawn. You can therefore skip the group = sex mapping. Using color has the added benefit that it uses color to distinguish the groups. Change the density plots from the previous exercise to add color.

## edit the next line to use color instead of group then add a density layer
heights %>% 
  ggplot(aes(height, color = sex)) + geom_density()

  1. We can also assign groups using the fill argument.

When using the geom_density geometry, color creates a colored line for the smooth density plot while fill colors in the area under the curve.

We can see what this looks like by running the following code:

heights %>% 
  ggplot(aes(height, fill = sex)) + 
  geom_density()

However, here the second density is drawn over the other. We can change this by using something called alpha blending.

heights %>% 
  ggplot(aes(height, fill = sex)) + 
  geom_density(alpha=0.2)

Section 3 Overview

Section 3 introduces you to summarizing with dplyr.

After completing Section 3, you will:

dplyr

The textbook for this section is available here

Key points

Code

# compute average and standard deviation for males
s <- heights %>%
    filter(sex == "Male") %>%
    summarize(average = mean(height), standard_deviation = sd(height))
    
# access average and standard deviation from summary table
s$average
## [1] 69.31475
s$standard_deviation
## [1] 3.611024
# compute median, min and max
heights %>%
    filter(sex == "Male") %>%
    summarize(median = median(height),
                       minimum = min(height),
                       maximum = max(height))
##   median minimum  maximum
## 1     69      50 82.67717
# alternative way to get min, median, max in base R
quantile(heights$height, c(0, 0.5, 1))
##       0%      50%     100% 
## 50.00000 68.50000 82.67717
# generates an error: summarize can only take functions that return a single value
heights %>%
    filter(sex == "Male") %>%
    summarize(range = quantile(height, c(0, 0.5, 1)))

The Dot Placeholder

The textbook for this section is available here

Note that a common replacement for the dot operator is the pull function. Here is the textbook section on the pull function.

Key points

Code

murders <- murders %>% mutate(murder_rate = total/population*100000)
summarize(murders, mean(murder_rate))
##   mean(murder_rate)
## 1          2.779125
# calculate US murder rate, generating a data frame
us_murder_rate <- murders %>%
    summarize(rate = sum(total) / sum(population) * 100000)
us_murder_rate
##       rate
## 1 3.034555
# extract the numeric US murder rate with the dot operator
us_murder_rate %>% .$rate
## [1] 3.034555
# calculate and extract the murder rate with one pipe
us_murder_rate <- murders %>%
    summarize(rate = sum(total) / sum(population * 100000)) %>%
    .$rate

Group By

The textbook for this section is available here

Key points

Code

# compute separate average and standard deviation for male/female heights
heights %>%
    group_by(sex) %>%
    summarize(average = mean(height), standard_deviation = sd(height))
## `summarise()` ungrouping output (override with `.groups` argument)
## # A tibble: 2 x 3
##   sex    average standard_deviation
##   <fct>    <dbl>              <dbl>
## 1 Female    64.9               3.76
## 2 Male      69.3               3.61
# compute median murder rate in 4 regions of country
murders <- murders %>%
    mutate(murder_rate = total/population * 100000)
murders %>%
    group_by(region) %>%
    summarize(median_rate = median(murder_rate))
## `summarise()` ungrouping output (override with `.groups` argument)
## # A tibble: 4 x 2
##   region        median_rate
##   <fct>               <dbl>
## 1 Northeast            1.80
## 2 South                3.40
## 3 North Central        1.97
## 4 West                 1.29

Sorting Data Tables

The textbook for this section is available here

Key points

Code

# set up murders object
murders <- murders %>%
    mutate(murder_rate = total/population * 100000)

# arrange by population column, smallest to largest
murders %>% arrange(population) %>% head()
##                  state abb        region population total murder_rate
## 1              Wyoming  WY          West     563626     5   0.8871131
## 2 District of Columbia  DC         South     601723    99  16.4527532
## 3              Vermont  VT     Northeast     625741     2   0.3196211
## 4         North Dakota  ND North Central     672591     4   0.5947151
## 5               Alaska  AK          West     710231    19   2.6751860
## 6         South Dakota  SD North Central     814180     8   0.9825837
# arrange by murder rate, smallest to largest
murders %>% arrange(murder_rate) %>% head()
##           state abb        region population total murder_rate
## 1       Vermont  VT     Northeast     625741     2   0.3196211
## 2 New Hampshire  NH     Northeast    1316470     5   0.3798036
## 3        Hawaii  HI          West    1360301     7   0.5145920
## 4  North Dakota  ND North Central     672591     4   0.5947151
## 5          Iowa  IA North Central    3046355    21   0.6893484
## 6         Idaho  ID          West    1567582    12   0.7655102
# arrange by murder rate in descending order
murders %>% arrange(desc(murder_rate)) %>% head()
##                  state abb        region population total murder_rate
## 1 District of Columbia  DC         South     601723    99   16.452753
## 2            Louisiana  LA         South    4533372   351    7.742581
## 3             Missouri  MO North Central    5988927   321    5.359892
## 4             Maryland  MD         South    5773552   293    5.074866
## 5       South Carolina  SC         South    4625364   207    4.475323
## 6             Delaware  DE         South     897934    38    4.231937
# arrange by region alphabetically, then by murder rate within each region
murders %>% arrange(region, murder_rate) %>% head()
##           state abb    region population total murder_rate
## 1       Vermont  VT Northeast     625741     2   0.3196211
## 2 New Hampshire  NH Northeast    1316470     5   0.3798036
## 3         Maine  ME Northeast    1328361    11   0.8280881
## 4  Rhode Island  RI Northeast    1052567    16   1.5200933
## 5 Massachusetts  MA Northeast    6547629   118   1.8021791
## 6      New York  NY Northeast   19378102   517   2.6679599
# show the top 10 states with highest murder rate, not ordered by rate
murders %>% top_n(10, murder_rate)
##                   state abb        region population total murder_rate
## 1               Arizona  AZ          West    6392017   232    3.629527
## 2              Delaware  DE         South     897934    38    4.231937
## 3  District of Columbia  DC         South     601723    99   16.452753
## 4               Georgia  GA         South    9920000   376    3.790323
## 5             Louisiana  LA         South    4533372   351    7.742581
## 6              Maryland  MD         South    5773552   293    5.074866
## 7              Michigan  MI North Central    9883640   413    4.178622
## 8           Mississippi  MS         South    2967297   120    4.044085
## 9              Missouri  MO North Central    5988927   321    5.359892
## 10       South Carolina  SC         South    4625364   207    4.475323
# show the top 10 states with highest murder rate, ordered by rate
murders %>% arrange(desc(murder_rate)) %>% top_n(10)
## Selecting by murder_rate
##                   state abb        region population total murder_rate
## 1  District of Columbia  DC         South     601723    99   16.452753
## 2             Louisiana  LA         South    4533372   351    7.742581
## 3              Missouri  MO North Central    5988927   321    5.359892
## 4              Maryland  MD         South    5773552   293    5.074866
## 5        South Carolina  SC         South    4625364   207    4.475323
## 6              Delaware  DE         South     897934    38    4.231937
## 7              Michigan  MI North Central    9883640   413    4.178622
## 8           Mississippi  MS         South    2967297   120    4.044085
## 9               Georgia  GA         South    9920000   376    3.790323
## 10              Arizona  AZ          West    6392017   232    3.629527

Assessment - Summarizing with dplyr

To practice our dplyr skills we will be working with data from the survey collected by the United States National Center for Health Statistics (NCHS). This center has conducted a series of health and nutrition surveys since the 1960’s.

Starting in 1999, about 5,000 individuals of all ages have been interviewed every year and then they complete the health examination component of the survey. Part of this dataset is made available via the NHANES package which can be loaded this way:

if(!require(NHANES)) install.packages("NHANES")
## Loading required package: NHANES
## Warning: package 'NHANES' was built under R version 4.0.2
library(NHANES)
data(NHANES)

The NHANES data has many missing values. Remember that the main summarization function in R will return NA if any of the entries of the input vector is an NA. Here is an example:

data(na_example)
mean(na_example)
## [1] NA
sd(na_example)
## [1] NA

To ignore the NAs, we can use the na.rm argument:

mean(na_example, na.rm = TRUE)
## [1] 2.301754
sd(na_example, na.rm = TRUE)
## [1] 1.22338

Try running this code, then let us know you are ready to proceed with the analysis.

  1. Let’s explore the NHANES data. We will be exploring blood pressure in this dataset.

First let’s select a group to set the standard. We will use 20-29 year old females. Note that the category is coded with 20-29, with a space in front of the 20! The AgeDecade is a categorical variable with these ages.

To know if someone is female, you can look at the Gender variable.

## fill in what is needed
tab <- NHANES %>% filter(AgeDecade == " 20-29" & Gender == "female")
head(tab)
## # A tibble: 6 x 76
##      ID SurveyYr Gender   Age AgeDecade AgeMonths Race1 Race3 Education
##   <int> <fct>    <fct>  <int> <fct>         <int> <fct> <fct> <fct>    
## 1 51710 2009_10  female    26 " 20-29"        319 White <NA>  College …
## 2 51731 2009_10  female    28 " 20-29"        346 Black <NA>  High Sch…
## 3 51741 2009_10  female    21 " 20-29"        253 Black <NA>  Some Col…
## 4 51741 2009_10  female    21 " 20-29"        253 Black <NA>  Some Col…
## 5 51760 2009_10  female    27 " 20-29"        334 Hisp… <NA>  9 - 11th…
## 6 51764 2009_10  female    29 " 20-29"        357 White <NA>  College …
## # … with 67 more variables: MaritalStatus <fct>, HHIncome <fct>,
## #   HHIncomeMid <int>, Poverty <dbl>, HomeRooms <int>, HomeOwn <fct>,
## #   Work <fct>, Weight <dbl>, Length <dbl>, HeadCirc <dbl>, Height <dbl>,
## #   BMI <dbl>, BMICatUnder20yrs <fct>, BMI_WHO <fct>, Pulse <int>,
## #   BPSysAve <int>, BPDiaAve <int>, BPSys1 <int>, BPDia1 <int>, BPSys2 <int>,
## #   BPDia2 <int>, BPSys3 <int>, BPDia3 <int>, Testosterone <dbl>,
## #   DirectChol <dbl>, TotChol <dbl>, UrineVol1 <int>, UrineFlow1 <dbl>,
## #   UrineVol2 <int>, UrineFlow2 <dbl>, Diabetes <fct>, DiabetesAge <int>,
## #   HealthGen <fct>, DaysPhysHlthBad <int>, DaysMentHlthBad <int>,
## #   LittleInterest <fct>, Depressed <fct>, nPregnancies <int>, nBabies <int>,
## #   Age1stBaby <int>, SleepHrsNight <int>, SleepTrouble <fct>,
## #   PhysActive <fct>, PhysActiveDays <int>, TVHrsDay <fct>, CompHrsDay <fct>,
## #   TVHrsDayChild <int>, CompHrsDayChild <int>, Alcohol12PlusYr <fct>,
## #   AlcoholDay <int>, AlcoholYear <int>, SmokeNow <fct>, Smoke100 <fct>,
## #   Smoke100n <fct>, SmokeAge <int>, Marijuana <fct>, AgeFirstMarij <int>,
## #   RegularMarij <fct>, AgeRegMarij <int>, HardDrugs <fct>, SexEver <fct>,
## #   SexAge <int>, SexNumPartnLife <int>, SexNumPartYear <int>, SameSex <fct>,
## #   SexOrientation <fct>, PregnantNow <fct>
  1. Now we will compute the average and standard deviation for the subgroup we defined in the previous exercise (20-29 year old females), which we will use reference for what is typical.

You will determine the average and standard deviation of systolic blood pressure, which are stored in the BPSysAve variable in the NHANES dataset.

## complete this line of code.
ref <- NHANES %>% filter(AgeDecade == " 20-29" & Gender == "female") %>% summarize(average = mean(BPSysAve, na.rm = TRUE), standard_deviation = sd(BPSysAve, na.rm = TRUE))
ref
## # A tibble: 1 x 2
##   average standard_deviation
##     <dbl>              <dbl>
## 1    108.               10.1
  1. Now we will repeat the exercise and generate only the average blood pressure for 20-29 year old females.

For this exercise, you should review how to use the place holder . in dplyr or the pull function.

## modify the code we wrote for previous exercise.
ref_avg <- NHANES %>%
  filter(AgeDecade == " 20-29" & Gender == "female") %>%
  summarize(average = mean(BPSysAve, na.rm = TRUE), 
            standard_deviation = sd(BPSysAve, na.rm=TRUE)) %>% .$average
ref_avg
## [1] 108.4224
  1. Let’s continue practicing by calculating two other data summaries: the minimum and the maximum.

Again we will do it for the BPSysAve variable and the group of 20-29 year old females.

## complete the line
NHANES %>%
      filter(AgeDecade == " 20-29"  & Gender == "female") %>% summarize(minbp = min(BPSysAve, na.rm = TRUE), 
            maxbp = max(BPSysAve, na.rm=TRUE))
## # A tibble: 1 x 2
##   minbp maxbp
##   <int> <int>
## 1    84   179
  1. Now let’s practice using the group_by function.

What we are about to do is a very common operation in data science: you will split a data table into groups and then compute summary statistics for each group.

We will compute the average and standard deviation of systolic blood pressure for females for each age group separately. Remember that the age groups are contained in AgeDecade.

##complete the line with group_by and summarize
NHANES %>%
      filter(Gender == "female") %>% group_by(AgeDecade) %>% summarize(average = mean(BPSysAve, na.rm = TRUE), 
            standard_deviation = sd(BPSysAve, na.rm=TRUE))
## `summarise()` ungrouping output (override with `.groups` argument)
## # A tibble: 9 x 3
##   AgeDecade average standard_deviation
##   <fct>       <dbl>              <dbl>
## 1 " 0-9"       100.               9.07
## 2 " 10-19"     104.               9.46
## 3 " 20-29"     108.              10.1 
## 4 " 30-39"     111.              12.3 
## 5 " 40-49"     115.              14.5 
## 6 " 50-59"     122.              16.2 
## 7 " 60-69"     127.              17.1 
## 8 " 70+"       134.              19.8 
## 9  <NA>        142.              22.9
  1. Now let’s practice using group_by some more.

We are going to repeat the previous exercise of calculating the average and standard deviation of systolic blood pressure, but for males instead of females.

This time we will not provide much sample code. You are on your own!

NHANES %>%
      filter(Gender == "male") %>% group_by(AgeDecade) %>% summarize(average = mean(BPSysAve, na.rm = TRUE), 
            standard_deviation = sd(BPSysAve, na.rm=TRUE))
## `summarise()` ungrouping output (override with `.groups` argument)
## # A tibble: 9 x 3
##   AgeDecade average standard_deviation
##   <fct>       <dbl>              <dbl>
## 1 " 0-9"       97.4               8.32
## 2 " 10-19"    110.               11.2 
## 3 " 20-29"    118.               11.3 
## 4 " 30-39"    119.               12.3 
## 5 " 40-49"    121.               14.0 
## 6 " 50-59"    126.               17.8 
## 7 " 60-69"    127.               17.5 
## 8 " 70+"      130.               18.7 
## 9  <NA>       136.               23.5
  1. We can actually combine both of these summaries into a single line of code.

This is because group_by permits us to group by more than one variable.

We can use group_by(AgeDecade, Gender) to group by both age decades and gender.

NHANES %>% group_by(AgeDecade, Gender) %>% summarize(average = mean(BPSysAve, na.rm = TRUE), 
            standard_deviation = sd(BPSysAve, na.rm=TRUE))
## `summarise()` regrouping output by 'AgeDecade' (override with `.groups` argument)
## # A tibble: 18 x 4
## # Groups:   AgeDecade [9]
##    AgeDecade Gender average standard_deviation
##    <fct>     <fct>    <dbl>              <dbl>
##  1 " 0-9"    female   100.                9.07
##  2 " 0-9"    male      97.4               8.32
##  3 " 10-19"  female   104.                9.46
##  4 " 10-19"  male     110.               11.2 
##  5 " 20-29"  female   108.               10.1 
##  6 " 20-29"  male     118.               11.3 
##  7 " 30-39"  female   111.               12.3 
##  8 " 30-39"  male     119.               12.3 
##  9 " 40-49"  female   115.               14.5 
## 10 " 40-49"  male     121.               14.0 
## 11 " 50-59"  female   122.               16.2 
## 12 " 50-59"  male     126.               17.8 
## 13 " 60-69"  female   127.               17.1 
## 14 " 60-69"  male     127.               17.5 
## 15 " 70+"    female   134.               19.8 
## 16 " 70+"    male     130.               18.7 
## 17  <NA>     female   142.               22.9 
## 18  <NA>     male     136.               23.5
  1. Now we are going to explore differences in systolic blood pressure across races, as reported in the Race1 variable.

We will learn to use the arrange function to order the outcome acording to one variable.

Note that this function can be used to order any table by a given outcome. Here is an example that arranges by systolic blood pressure.

NHANES %>% arrange(BPSysAve)

If we want it in descending order we can use the desc function like this:

NHANES %>% arrange(desc(BPSysAve))

In this example, we will compare systolic blood pressure across values of the Race1 variable for males between the ages of 40-49.

NHANES %>% filter(AgeDecade == " 40-49" & Gender == "male") %>% group_by(Race1) %>% summarize(average = mean(BPSysAve, na.rm = TRUE), standard_deviation = sd(BPSysAve, na.rm=TRUE)) %>% arrange(average)
## `summarise()` ungrouping output (override with `.groups` argument)
## # A tibble: 5 x 3
##   Race1    average standard_deviation
##   <fct>      <dbl>              <dbl>
## 1 White       120.               13.4
## 2 Other       120.               16.2
## 3 Hispanic    122.               11.1
## 4 Mexican     122.               13.9
## 5 Black       126.               17.1

Section 4 Overview

In Section 4, you will look at a case study involving data from the Gapminder Foundation about trends in world health and economics.

After completing Section 4, you will:

Gapminder Dataset

The textbook for this section is available here

Key points

Code

# load and inspect gapminder data
data(gapminder)
head(gapminder)
##               country year infant_mortality life_expectancy fertility
## 1             Albania 1960           115.40           62.87      6.19
## 2             Algeria 1960           148.20           47.50      7.65
## 3              Angola 1960           208.00           35.98      7.32
## 4 Antigua and Barbuda 1960               NA           62.97      4.43
## 5           Argentina 1960            59.87           65.39      3.11
## 6             Armenia 1960               NA           66.86      4.55
##   population          gdp continent          region
## 1    1636054           NA    Europe Southern Europe
## 2   11124892  13828152297    Africa Northern Africa
## 3    5270844           NA    Africa   Middle Africa
## 4      54681           NA  Americas       Caribbean
## 5   20619075 108322326649  Americas   South America
## 6    1867396           NA      Asia    Western Asia
# compare infant mortality in Sri Lanka and Turkey
gapminder %>%
    filter(year == 2015 & country %in% c("Sri Lanka", "Turkey")) %>%
    select(country, infant_mortality)
##     country infant_mortality
## 1 Sri Lanka              8.4
## 2    Turkey             11.6

Life Expectancy and Fertility Rates

The textbook for this section is available here

Key points

Code

# basic scatterplot of life expectancy versus fertility
ds_theme_set()    # set plot theme
filter(gapminder, year == 1962) %>%
    ggplot(aes(fertility, life_expectancy)) +
    geom_point()

# add color as continent
filter(gapminder, year == 1962) %>%
    ggplot(aes(fertility, life_expectancy, color = continent)) +
    geom_point()

Faceting

The textbook for this section is available here

Key points

Code

# facet by continent and year
filter(gapminder, year %in% c(1962, 2012)) %>%
    ggplot(aes(fertility, life_expectancy, col = continent)) +
    geom_point() +
    facet_grid(continent ~ year)

# facet by year only
filter(gapminder, year %in% c(1962, 2012)) %>%
    ggplot(aes(fertility, life_expectancy, col = continent)) +
    geom_point() +
    facet_grid(. ~ year)

# facet by year, plots wrapped onto multiple rows
years <- c(1962, 1980, 1990, 2000, 2012)
continents <- c("Europe", "Asia")
gapminder %>%
    filter(year %in% years & continent %in% continents) %>%
    ggplot(aes(fertility, life_expectancy, col = continent)) +
    geom_point() +
    facet_wrap(~year)

Time Series Plots

The textbook for this section is available here

Key points

Code: Single time series

# scatterplot of US fertility by year
gapminder %>%
    filter(country == "United States") %>%
    ggplot(aes(year, fertility)) +
    geom_point()
## Warning: Removed 1 rows containing missing values (geom_point).

# line plot of US fertility by year
gapminder %>%
    filter(country == "United States") %>%
    ggplot(aes(year, fertility)) +
    geom_line()
## Warning: Removed 1 row(s) containing missing values (geom_path).

Code: Multiple time series

# line plot fertility time series for two countries- only one line (incorrect)
countries <- c("South Korea", "Germany")
gapminder %>% filter(country %in% countries) %>%
    ggplot(aes(year, fertility)) +
    geom_line()
## Warning: Removed 2 row(s) containing missing values (geom_path).

# line plot fertility time series for two countries - one line per country
gapminder %>% filter(country %in% countries) %>%
    ggplot(aes(year, fertility, group = country)) +
    geom_line()
## Warning: Removed 2 row(s) containing missing values (geom_path).

# fertility time series for two countries - lines colored by country
gapminder %>% filter(country %in% countries) %>%
    ggplot(aes(year, fertility, col = country)) +
    geom_line()
## Warning: Removed 2 row(s) containing missing values (geom_path).

Code: Adding text labels to a plot

# life expectancy time series - lines colored by country and labeled, no legend
labels <- data.frame(country = countries, x = c(1975, 1965), y = c(60, 72))
gapminder %>% filter(country %in% countries) %>%
    ggplot(aes(year, life_expectancy, col = country)) +
    geom_line() +
    geom_text(data = labels, aes(x, y, label = country), size = 5) +
    theme(legend.position = "none")

Transformations

The textbook for this section is available here and [here(https://rafalab.github.io/dsbook/gapminder.html#visualizing-multimodal-distributions)

Key points

Code

# add dollars per day variable
gapminder <- gapminder %>%
    mutate(dollars_per_day = gdp/population/365)

# histogram of dollars per day
past_year <- 1970
gapminder %>%
    filter(year == past_year & !is.na(gdp)) %>%
    ggplot(aes(dollars_per_day)) +
    geom_histogram(binwidth = 1, color = "black")

# repeat histogram with log2 scaled data
gapminder %>%
    filter(year == past_year & !is.na(gdp)) %>%
    ggplot(aes(log2(dollars_per_day))) +
    geom_histogram(binwidth = 1, color = "black")

# repeat histogram with log2 scaled x-axis
gapminder %>%
    filter(year == past_year & !is.na(gdp)) %>%
    ggplot(aes(dollars_per_day)) +
    geom_histogram(binwidth = 1, color = "black") +
    scale_x_continuous(trans = "log2")

Stratify and Boxplot

The textbook for this section is available here. Note that many boxplots from the video are instead dot plots in the textbook and that a different boxplot is constructed in the textbook. Also read that section to see an example of grouping factors with the case_when function.

Key points

Code: Boxplot of GDP by region

# add dollars per day variable
gapminder <- gapminder %>%
    mutate(dollars_per_day = gdp/population/365)

# number of regions
length(levels(gapminder$region))
## [1] 22
# boxplot of GDP by region in 1970
past_year <- 1970
p <- gapminder %>%
    filter(year == past_year & !is.na(gdp)) %>%
    ggplot(aes(region, dollars_per_day))
p + geom_boxplot()

# rotate names on x-axis
p + geom_boxplot() +
    theme(axis.text.x = element_text(angle = 90, hjust = 1))

Code: The reorder function

# by default, factor order is alphabetical
fac <- factor(c("Asia", "Asia", "West", "West", "West"))
levels(fac)
## [1] "Asia" "West"
# reorder factor by the category means
value <- c(10, 11, 12, 6, 4)
fac <- reorder(fac, value, FUN = mean)
levels(fac)
## [1] "West" "Asia"

Code: Enhanced boxplot ordered by median income, scaled, and showing data

# reorder by median income and color by continent
p <- gapminder %>%
    filter(year == past_year & !is.na(gdp)) %>%
    mutate(region = reorder(region, dollars_per_day, FUN = median)) %>%    # reorder
    ggplot(aes(region, dollars_per_day, fill = continent)) +    # color by continent
    geom_boxplot() +
    theme(axis.text.x = element_text(angle = 90, hjust = 1)) +
    xlab("")
p

# log2 scale y-axis
p + scale_y_continuous(trans = "log2")

# add data points
p + scale_y_continuous(trans = "log2") + geom_point(show.legend = FALSE)

Comparing Distributions

The textbook for this section is available here. Note that the boxplots are slightly different.

Key points

Code: Histogram of income in West versus developing world, 1970 and 2010

# add dollars per day variable and define past year
gapminder <- gapminder %>%
    mutate(dollars_per_day = gdp/population/365)
past_year <- 1970

# define Western countries
west <- c("Western Europe", "Northern Europe", "Southern Europe", "Northern America", "Australia and New Zealand")

# facet by West vs devloping
gapminder %>%
    filter(year == past_year & !is.na(gdp)) %>%
    mutate(group = ifelse(region %in% west, "West", "Developing")) %>%
    ggplot(aes(dollars_per_day)) +
    geom_histogram(binwidth = 1, color = "black") +
    scale_x_continuous(trans = "log2") +
    facet_grid(. ~ group)

# facet by West/developing and year
present_year <- 2010
gapminder %>%
    filter(year %in% c(past_year, present_year) & !is.na(gdp)) %>%
    mutate(group = ifelse(region %in% west, "West", "Developing")) %>%
    ggplot(aes(dollars_per_day)) +
    geom_histogram(binwidth = 1, color = "black") +
    scale_x_continuous(trans = "log2") +
    facet_grid(year ~ group)

Code: Income distribution of West versus developing world, only countries with data

# define countries that have data available in both years
country_list_1 <- gapminder %>%
    filter(year == past_year & !is.na(dollars_per_day)) %>% .$country
country_list_2 <- gapminder %>%
    filter(year == present_year & !is.na(dollars_per_day)) %>% .$country
country_list <- intersect(country_list_1, country_list_2)

# make histogram including only countries with data available in both years
gapminder %>%
    filter(year %in% c(past_year, present_year) & country %in% country_list) %>%    # keep only selected countries
    mutate(group = ifelse(region %in% west, "West", "Developing")) %>%
    ggplot(aes(dollars_per_day)) +
    geom_histogram(binwidth = 1, color = "black") +
    scale_x_continuous(trans = "log2") +
    facet_grid(year ~ group)

Code: Boxplots of income in West versus developing world, 1970 and 2010

p <- gapminder %>%
    filter(year %in% c(past_year, present_year) & country %in% country_list) %>%
    mutate(region = reorder(region, dollars_per_day, FUN = median)) %>%
    ggplot() +
    theme(axis.text.x = element_text(angle = 90, hjust = 1)) +
    xlab("") + scale_y_continuous(trans = "log2")
p + geom_boxplot(aes(region, dollars_per_day, fill = continent)) +
    facet_grid(year ~ .)

# arrange matching boxplots next to each other, colored by year
p + geom_boxplot(aes(region, dollars_per_day, fill = factor(year)))

Density Plots

The textbook for this section is available:

Key points

Code: Faceted smooth density plots

# smooth density plots - area under each curve adds to 1
gapminder %>%
    filter(year == past_year & country %in% country_list) %>%
    mutate(group = ifelse(region %in% west, "West", "Developing")) %>% group_by(group) %>%
    summarize(n = n()) %>% knitr::kable()
## `summarise()` ungrouping output (override with `.groups` argument)
group n
Developing 87
West 21 
# smooth density plots - variable counts on y-axis
p <- gapminder %>%
    filter(year == past_year & country %in% country_list) %>%
    mutate(group = ifelse(region %in% west, "West", "Developing")) %>%
    ggplot(aes(dollars_per_day, y = ..count.., fill = group)) +
    scale_x_continuous(trans = "log2")
p + geom_density(alpha = 0.2, bw = 0.75) + facet_grid(year ~ .)

Code: Add new region groups with case_when

# add group as a factor, grouping regions
gapminder <- gapminder %>%
    mutate(group = case_when(
        .$region %in% west ~ "West",
        .$region %in% c("Eastern Asia", "South-Eastern Asia") ~ "East Asia",
        .$region %in% c("Caribbean", "Central America", "South America") ~ "Latin America",
        .$continent == "Africa" & .$region != "Northern Africa" ~ "Sub-Saharan Africa",
        TRUE ~ "Others"))

# reorder factor levels
gapminder <- gapminder %>%
    mutate(group = factor(group, levels = c("Others", "Latin America", "East Asia", "Sub-Saharan Africa", "West")))

Code: Stacked density plot

# note you must redefine p with the new gapminder object first
p <- gapminder %>%
  filter(year %in% c(past_year, present_year) & country %in% country_list) %>%
    ggplot(aes(dollars_per_day, fill = group)) +
    scale_x_continuous(trans = "log2")

# stacked density plot
p + geom_density(alpha = 0.2, bw = 0.75, position = "stack") +
    facet_grid(year ~ .)

Code: Weighted stacked density plot

# weighted stacked density plot
gapminder %>%
    filter(year %in% c(past_year, present_year) & country %in% country_list) %>%
    group_by(year) %>%
    mutate(weight = population/sum(population*2)) %>%
    ungroup() %>%
    ggplot(aes(dollars_per_day, fill = group, weight = weight)) +
    scale_x_continuous(trans = "log2") +
    geom_density(alpha = 0.2, bw = 0.75, position = "stack") + facet_grid(year ~ .)

Ecological Fallacy

The textbook for this section is available here

Key points

Code

# add additional cases
gapminder <- gapminder %>%
    mutate(group = case_when(
        .$region %in% west ~ "The West",
        .$region %in% "Northern Africa" ~ "Northern Africa",
        .$region %in% c("Eastern Asia", "South-Eastern Asia") ~ "East Asia",
        .$region == "Southern Asia" ~ "Southern Asia",
        .$region %in% c("Central America", "South America", "Caribbean") ~ "Latin America",
        .$continent == "Africa" & .$region != "Northern Africa" ~ "Sub-Saharan Africa",
        .$region %in% c("Melanesia", "Micronesia", "Polynesia") ~ "Pacific Islands"))

# define a data frame with group average income and average infant survival rate
surv_income <- gapminder %>%
    filter(year %in% present_year & !is.na(gdp) & !is.na(infant_mortality) & !is.na(group)) %>%
    group_by(group) %>%
    summarize(income = sum(gdp)/sum(population)/365,
                        infant_survival_rate = 1 - sum(infant_mortality/1000*population)/sum(population))
## `summarise()` ungrouping output (override with `.groups` argument)
surv_income %>% arrange(income)
## # A tibble: 7 x 3
##   group              income infant_survival_rate
##   <chr>               <dbl>                <dbl>
## 1 Sub-Saharan Africa   1.76                0.936
## 2 Southern Asia        2.07                0.952
## 3 Pacific Islands      2.70                0.956
## 4 Northern Africa      4.94                0.970
## 5 Latin America       13.2                 0.983
## 6 East Asia           13.4                 0.985
## 7 The West            77.1                 0.995
# plot infant survival versus income, with transformed axes
surv_income %>% ggplot(aes(income, infant_survival_rate, label = group, color = group)) +
    scale_x_continuous(trans = "log2", limit = c(0.25, 150)) +
    scale_y_continuous(trans = "logit", limit = c(0.875, .9981),
                                       breaks = c(.85, .90, .95, .99, .995, .998)) +
    geom_label(size = 3, show.legend = FALSE)

Assessment 8 (Exploring the Gapminder Dataset)

  1. Life expectancy vs fertility - part 1

The Gapminder Foundation is a non-profit organization based in Sweden that promotes global development through the use of statistics that can help reduce misconceptions about global development.

library(dplyr)
library(ggplot2)
library(dslabs)
data(gapminder)
head(gapminder)
   country             year       infant_mortality      life_expectancy     fertility     population
   <fctr>              <int>      <dbl>                 <dbl>               <dbl>         <dbl>
1  Albania         1960   115.40            62.87               6.19      1636054   
2  Algeria         1960   148.20            47.50               7.65      11124892  
3  Angola          1960   208.00            35.98               7.32      5270844   
4  Antigua and Barbuda 1960   NA                    62.97               4.43      54681 
5  Argentina           1960   59.87                 65.39               3.11      20619075  
6  Armenia         1960   NA                    66.86               4.55      1867396   
6 rows | 1-7 of 10 columns
## fill out the missing parts in filter and aes
gapminder %>% filter(continent=="Africa" & year=="2012") %>%
  ggplot(aes(fertility,life_expectancy)) +
  geom_point()

index

  1. Life expectancy vs fertility - part 2 - coloring your plot

Note that there is quite a bit of variability in life expectancy and fertility with some African countries having very high life expectancies. There also appear to be three clusters in the plot.

Remake the plot from the previous exercises but this time use color to dinstinguish the different regions of Africa to see if this explains the clusters. Remember that you can explore the gapminder data to see how the regions of Africa are labeled in the dataframe!

library(dplyr)
library(ggplot2)
library(dslabs)
data(gapminder)

gapminder %>% filter(continent=="Africa" & year=="2012") %>%
  ggplot(aes(fertility,life_expectancy, color=region)) +
  geom_point()

index

  1. Life expectancy vs fertility - part 3 - selecting country and region

While many of the countries in the high life expectancy/low fertility cluster are from Northern Africa, three countries are not.

library(dplyr)
library(dslabs)
data(gapminder)
head(gapminder)
   country             year       infant_mortality      life_expectancy     fertility     population
   <fctr>              <int>      <dbl>                 <dbl>               <dbl>         <dbl>
1  Albania         1960   115.40            62.87               6.19      1636054   
2  Algeria         1960   148.20            47.50               7.65      11124892  
3  Angola          1960   208.00            35.98               7.32      5270844   
4  Antigua and Barbuda 1960   NA                    62.97               4.43      54681 
5  Argentina           1960   59.87                 65.39               3.11      20619075  
6  Armenia         1960   NA                    66.86               4.55      1867396   
6 rows | 1-7 of 10 columns
df <- gapminder %>% filter(continent=="Africa" & year=="2012" & fertility <=3 & life_expectancy>=70) %>%
  select(country,region)
  1. Life expectancy and the Vietnam War - part 1

The Vietnam War lasted from 1955 to 1975. Do the data support war having a negative effect on life expectancy? We will create a time series plot that covers the period from 1960 to 2010 of life expectancy for Vietnam and the United States, using color to distinguish the two countries. In this start we start the analysis by generating a table.

library(dplyr)
library(dslabs)
data(gapminder)

head(gapminder)
   country             year       infant_mortality      life_expectancy     fertility     population
   <fctr>              <int>      <dbl>                 <dbl>               <dbl>         <dbl>
1  Albania         1960   115.40            62.87               6.19      1636054   
2  Algeria         1960   148.20            47.50               7.65      11124892  
3  Angola          1960   208.00            35.98               7.32      5270844   
4  Antigua and Barbuda 1960   NA                    62.97               4.43      54681 
5  Argentina           1960   59.87                 65.39               3.11      20619075  
6  Armenia         1960   NA                    66.86               4.55      1867396   
6 rows | 1-7 of 10 columns
tab <- gapminder %>% filter(year>=1960 & year<=2010 & country%in%c("Vietnam","United States"))
  1. Life expectancy and the Vietnam War - part 2

Now that you have created the data table in Exercise 4, it is time to plot the data for the two countries.

p <- tab %>% ggplot(aes(year,life_expectancy,color=country)) + geom_line()
p

index

  1. Life expectancy in Cambodia

Cambodia was also involved in this conflict and, after the war, Pol Pot and his communist Khmer Rouge took control and ruled Cambodia from 1975 to 1979. He is considered one of the most brutal dictators in history. Do the data support this claim?

Use a single line of code to create a time series plot from 1960 to 2010 of life expectancy vs year for Cambodia.

library(dplyr)
library(ggplot2)
library(dslabs)
data(gapminder)

gapminder %>% filter(year>=1960 & year <= 2010 & country=="Cambodia") %>% ggplot(aes(year,life_expectancy)) + geom_line()

index

  1. Dollars per day - part 1

Now we are going to calculate and plot dollars per day for African countries in 2010 using GDP data.

In the first part of this analysis, we will create the dollars per day variable. - Use mutate to create a dollars_per_day variable, which is defined as gdp/population/365. - Create the dollars_per_day variable for African countries for the year 2010. - Remove any NA values. - Save the mutated dataset as daydollars.

library(dplyr)
library(dslabs)
data(gapminder)
daydollars <- gapminder %>% mutate(dollars_per_day=gdp/population/365)%>% filter(year==2010 & continent=="Africa" & !is.na(dollars_per_day))
  1. Dollars per day - part 2

Now we are going to calculate and plot dollars per day for African countries in 2010 using GDP data.

In the second part of this analysis, we will plot the smooth density plot using a log (base 2) x axis. - The dataset including the dollars_per_day variable is preloaded as daydollars. - Create a smooth density plot of dollars per day from daydollars. - Use a log (base 2) scale for the x axis.

daydollars %>% ggplot(aes(dollars_per_day)) + geom_density() + scale_x_continuous(trans="log2")

index

  1. Dollars per day - part 3 - multiple density plots

Now we are going to combine the plotting tools we have used in the past two exercises to create density plots for multiple years. - Create the dollars_per_day variable as in Exercise 7, but for African countries in the years 1970 and 2010 this time. - Make sure you remove any NA values. - Create a smooth density plot of dollars per day for 1970 and 2010 using a log (base 2) scale for the x axis. - Use facet_grid to show a different density plot for 1970 and 2010.

library(dplyr)
library(ggplot2)
library(dslabs)
data(gapminder)

daydollars <- gapminder %>% mutate(dollars_per_day=gdp/population/365)%>% filter(year %in% c(1970,2010) & continent=="Africa" & !is.na(dollars_per_day))

daydollars %>% ggplot(aes(dollars_per_day)) + geom_density() + scale_x_continuous(trans='log2') + facet_grid(.~year)

index

  1. Dollars per day - part 4 - stacked histograms

Now we are going to edit the code from Exercise 9 to show stacked histograms of each region in Africa.

Much of the code will be the same as in Exercise 9: - Create the dollars_per_day variable as in Exercise 7, but for African countries in the years 1970 and 2010 this time. - Make sure you remove any NA values. - Create a smooth density plot of dollars per day for 1970 and 2010 using a log (base 2) scale for the x axis. - Use facet_grid to show a different density plot for 1970 and 2010. - Make sure the densities are smooth by using bw = 0.5. - Use the fill and position arguments where appropriate to create the stacked histograms of each region.

library(dplyr)
library(ggplot2)
library(dslabs)
data(gapminder)

daydollars <- gapminder %>% mutate(dollars_per_day=gdp/population/365)%>% filter(year %in% c(1970,2010) & continent=="Africa" & !is.na(dollars_per_day))

daydollars %>% ggplot(aes(dollars_per_day,fill = region)) + geom_density(bw=0.5,position='stack') + scale_x_continuous(trans='log2') + facet_grid(.~year)

index

  1. Infant mortality scatter plot - part 1

We are going to continue looking at patterns in the gapminder dataset by plotting infant mortality rates versus dollars per day for African countries. - Generate dollars_per_day using mutate and filter for the year 2010 for African countries. - Remember to remove NA values. - Store the mutated dataset in gapminder_Africa_2010. - Make a scatter plot of infant_mortaility versus dollars_per_day for countries in the African continent. - Use color to denote the different regions of Africa.

library(dplyr)
library(ggplot2)
library(dslabs)
data(gapminder)
gapminder_Africa_2010 <- daydollars <- gapminder %>% mutate(dollars_per_day=gdp/population/365)%>% filter(year %in% c(2010) & continent=="Africa" & !is.na(dollars_per_day))

# now make the scatter plot

gapminder_Africa_2010 %>% ggplot(aes(dollars_per_day,infant_mortality,color = region)) + geom_point()

index

  1. Infant mortality scatter plot - part 2 - logarithmic axis

Now we are going to transform the x axis of the plot from the previous exercise. - The mutated dataset is preloaded as gapminder_Africa_2010. - As in the previous exercise, make a scatter plot of infant_mortaility versus dollars_per_day for countries in the African continent. - As in the previous exercise, use color to denote the different regions of Africa. - Transform the x axis to be in the log (base 2) scale.

gapminder_Africa_2010 %>% ggplot(aes(dollars_per_day,infant_mortality, color=region)) + geom_point() + scale_x_continuous(trans='log2')

index

  1. Infant mortality scatter plot - part 3 - adding labels

Note that there is a large variation in infant mortality and dollars per day among African countries.

As an example, one country has infant mortality rates of less than 20 per 1000 and dollars per day of 16, while another country has infant mortality rates over 10% and dollars per day of about 1.

In this exercise, we will remake the plot from Exercise 12 with country names instead of points so we can identify which countries are which. - The mutated dataset is preloaded as gapminder_Africa_2010. - As in the previous exercise, make a scatter plot of infant_mortaility versus dollars_per_day for countries in the African continent. - As in the previous exercise, use color to denote the different regions of Africa. - As in the previous exercise, transform the x axis to be in the log (base 2) scale. - Add a layer to display country names instead of points.

gapminder_Africa_2010 %>% ggplot(aes(dollars_per_day,infant_mortality, color=region,label = country)) + geom_point() + scale_x_continuous(trans='log2') +
    geom_text()

index

  1. Infant mortality scatter plot - part 4 - comparison of scatter plots

Now we are going to look at changes in the infant mortality and dollars per day patterns African countries between 1970 and 2010. - Generate dollars_per_day using mutate and filter for the years 1970 and 2010 for African countries. - Remember to remove NA values. - As in the previous exercise, make a scatter plot of infant_mortaility versus dollars_per_day for countries in the African continent. - As in the previous exercise, use color to denote the different regions of Africa. - As in the previous exercise, transform the x axis to be in the log (base 2) scale. - As in the previous exercise, add a layer to display country names instead of points. - Use facet_grid to show different plots for 1970 and 2010.

library(dplyr)
library(ggplot2)
library(dslabs)
data(gapminder)
gapminder  %>%
    filter(continent == "Africa" & year %in% c(1970,2010) & !is.na(gdp) & !is.na(year) & !is.na(infant_mortality)) %>%
    mutate(dollars_per_day = gdp/population/365) %>%
    ggplot(aes(dollars_per_day, infant_mortality, color = region,label = country)) +
    geom_point() +
    scale_x_continuous(trans = "log2") +
    geom_text() +
    facet_grid(year~.)

index

Section 5 Overview

Section 5 covers some general principles that can serve as guides for effective data visualization.

After completing Section 5, you will: - understand basic principles of effective data visualization. - understand the importance of keeping your goal in mind when deciding on a visualization approach. - understand principles for encoding data, including position, aligned lengths, angles, area, brightness, and color hue. - know when to include the number zero in visualizations. - be able to use techniques to ease comparisons, such as using common axes, putting visual cues to be compared adjacent to one another, and using color effectively.

The textbook for this section is available here

Assessment 9 (Data Visualization Principles, Part 1)

1: Customizing plots - Pie charts

Pie charts are appropriate: - [ ] A. When we want to display percentages. - [ ] B. When ggplot2 is not available. - [ ] C. When I am in a bakery. - [X] D. Never. Barplots and tables are always better.

  1. Customizing plots - What’s wrong?

What is the problem with this plot?

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  1. Customizing plots - What’s wrong 2?.

Take a look at the following two plots. They show the same information: rates of measles by state in the United States for 1928.

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Assessment 10 (Data Visualization Principles, Part 2)

1: Customizing plots - watch and learn

To make the plot on the right in the exercise from the last set of assessments, we had to reorder the levels of the states’ variables. - Redefine the state object so that the levels are re-ordered by rate. - Print the new object state and its levels so you can see that the vector is now re-ordered by the levels.

library(dplyr)
library(ggplot2)
library(dslabs)
dat <- us_contagious_diseases %>%
filter(year == 1967 & disease=="Measles" & !is.na(population)) %>% mutate(rate = count / population * 10000 * 52 / weeks_reporting)
state <- dat$state 
rate <- dat$count/(dat$population/10000)*(52/dat$weeks_reporting)

state <- reorder(state,rate)
print(state)
##  [1] Alabama              Alaska               Arizona             
##  [4] Arkansas             California           Colorado            
##  [7] Connecticut          Delaware             District Of Columbia
## [10] Florida              Georgia              Hawaii              
## [13] Idaho                Illinois             Indiana             
## [16] Iowa                 Kansas               Kentucky            
## [19] Louisiana            Maine                Maryland            
## [22] Massachusetts        Michigan             Minnesota           
## [25] Mississippi          Missouri             Montana             
## [28] Nebraska             Nevada               New Hampshire       
## [31] New Jersey           New Mexico           New York            
## [34] North Carolina       North Dakota         Ohio                
## [37] Oklahoma             Oregon               Pennsylvania        
## [40] Rhode Island         South Carolina       South Dakota        
## [43] Tennessee            Texas                Utah                
## [46] Vermont              Virginia             Washington          
## [49] West Virginia        Wisconsin            Wyoming             
## attr(,"scores")
##              Alabama               Alaska              Arizona 
##           4.16107582           5.46389893           6.32695891 
##             Arkansas           California             Colorado 
##           6.87899954           2.79313560           7.96331905 
##          Connecticut             Delaware District Of Columbia 
##           0.36986840           1.13098183           0.35873614 
##              Florida              Georgia               Hawaii 
##           2.89358806           0.09987991           2.50173748 
##                Idaho             Illinois              Indiana 
##           6.03115170           1.20115480           1.34027323 
##                 Iowa               Kansas             Kentucky 
##           2.94948911           0.66386422           4.74576011 
##            Louisiana                Maine             Maryland 
##           0.46088071           2.57520433           0.49922233 
##        Massachusetts             Michigan            Minnesota 
##           0.74762338           1.33466700           0.37722410 
##          Mississippi             Missouri              Montana 
##           3.11366532           0.75696354           5.00433320 
##             Nebraska               Nevada        New Hampshire 
##           3.64389801           6.43683882           0.47181511 
##           New Jersey           New Mexico             New York 
##           0.88414264           6.15969926           0.66849058 
##       North Carolina         North Dakota                 Ohio 
##           1.92529764          14.48024642           1.16382241 
##             Oklahoma               Oregon         Pennsylvania 
##           3.27496900           8.75036439           0.67687303 
##         Rhode Island       South Carolina         South Dakota 
##           0.68207448           2.10412531           0.90289534 
##            Tennessee                Texas                 Utah 
##           5.47344506          12.49773953           4.03005836 
##              Vermont             Virginia           Washington 
##           1.00970314           5.28270939          17.65180349 
##        West Virginia            Wisconsin              Wyoming 
##           8.59456463           4.96246019           6.97303449 
## 51 Levels: Georgia District Of Columbia Connecticut ... Washington
levels(state)
##  [1] "Georgia"              "District Of Columbia" "Connecticut"         
##  [4] "Minnesota"            "Louisiana"            "New Hampshire"       
##  [7] "Maryland"             "Kansas"               "New York"            
## [10] "Pennsylvania"         "Rhode Island"         "Massachusetts"       
## [13] "Missouri"             "New Jersey"           "South Dakota"        
## [16] "Vermont"              "Delaware"             "Ohio"                
## [19] "Illinois"             "Michigan"             "Indiana"             
## [22] "North Carolina"       "South Carolina"       "Hawaii"              
## [25] "Maine"                "California"           "Florida"             
## [28] "Iowa"                 "Mississippi"          "Oklahoma"            
## [31] "Nebraska"             "Utah"                 "Alabama"             
## [34] "Kentucky"             "Wisconsin"            "Montana"             
## [37] "Virginia"             "Alaska"               "Tennessee"           
## [40] "Idaho"                "New Mexico"           "Arizona"             
## [43] "Nevada"               "Arkansas"             "Wyoming"             
## [46] "Colorado"             "West Virginia"        "Oregon"              
## [49] "Texas"                "North Dakota"         "Washington"
  1. Customizing plots - redefining

Now we are going to customize this plot a little more by creating a rate variable and reordering by that variable instead. - Add a single line of code to the definition of the dat table that uses mutate to reorder the states by the rate variable. - The sample code provided will then create a bar plot using the newly defined dat.

library(dplyr)
library(ggplot2)
library(dslabs)
data(us_contagious_diseases)
dat <- us_contagious_diseases %>% filter(year == 1967 & disease=="Measles" & count>0 & !is.na(population)) %>%
  mutate(rate = count / population * 10000 * 52 / weeks_reporting) %>% mutate(state = reorder(state, rate))
dat %>% ggplot(aes(state, rate)) +
  geom_bar(stat="identity") +
  coord_flip()

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  1. Showing the data and customizing plots

Say we are interested in comparing gun homicide rates across regions of the US. We see this plot:

library(dplyr)
library(ggplot2)
library(dslabs)
data("murders")
murders %>% mutate(rate = total/population*100000) %>%
  group_by(region) %>%
  summarize(avg = mean(rate)) %>%
  mutate(region = factor(region)) %>%
  ggplot(aes(region, avg)) +
  geom_bar(stat="identity") +
  ylab("Murder Rate Average")

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and decide to move to a state in the western region. What is the main problem with this interpretaion? - [ ] A. The categories are ordered alphabetically. - [ ] B. The graph does not show standard errors. - [X] C. It does not show all the data. We do not see the variability within a region and it’s possible that the safest states are not in the West. - [ ] D. The Northeast has the lowest average.

  1. Making a box plot

To further investigate whether moving to the western region is a wise decision, let’s make a box plot of murder rates by region, showing all points. - Make a box plot of the murder rates by region. - Order the regions by their median murder rate. - Show all of the points on the box plot.

library(dplyr)
library(ggplot2)
library(dslabs)
data("murders")
murders %>% mutate(rate = total/population*100000) %>%
  mutate(region=reorder(region, rate, FUN=median)) %>%
  ggplot(aes(region, rate)) +
  geom_boxplot() +
  geom_point()

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Assessment 11 (Data Visualization Principles, Part 3)

  1. Tile plot - measles and smallpox

The sample code given creates a tile plot showing the rate of measles cases per population. We are going to modify the tile plot to look at smallpox cases instead.

if(!require(RColorBrewer)) install.packages("RColorBrewer")

library(dplyr)
library(ggplot2)
library(RColorBrewer)
library(dslabs)
data(us_contagious_diseases)
head(us_contagious_diseases)
   disease       state      year      weeks_reporting     count      population
   <fctr>        <fctr>     <dbl>     <int>               <dbl>      <dbl>
1  Hepatitis A   Alabama    1966      50              321        3345787
2  Hepatitis A   Alabama    1967      49              291        3364130
3  Hepatitis A   Alabama    1968      52              314        3386068
4  Hepatitis A   Alabama    1969      49              380        3412450
5  Hepatitis A   Alabama    1970      51                  413        3444165
6  Hepatitis A   Alabama    1971      51              378        3481798
6 rows
the_disease = "Measles"
dat <- us_contagious_diseases %>% 
   filter(!state%in%c("Hawaii","Alaska") & disease == the_disease) %>% 
   mutate(rate = count / population * 10000) %>% 
   mutate(state = reorder(state, rate))

dat %>% ggplot(aes(year, state, fill = rate)) + 
  geom_tile(color = "grey50") + 
  scale_x_continuous(expand=c(0,0)) + 
  scale_fill_gradientn(colors = brewer.pal(9, "Reds"), trans = "sqrt") + 
  theme_minimal() + 
  theme(panel.grid = element_blank()) + 
  ggtitle(the_disease) + 
  ylab("") + 
  xlab("")

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library(dplyr)
library(ggplot2)
library(RColorBrewer)
library(dslabs)
data(us_contagious_diseases)
head(us_contagious_diseases)
   disease       state      year      weeks_reporting     count      population
   <fctr>        <fctr>     <dbl>     <int>               <dbl>      <dbl>
1  Hepatitis A   Alabama    1966      50              321        3345787
2  Hepatitis A   Alabama    1967      49              291        3364130
3  Hepatitis A   Alabama    1968      52              314        3386068
4  Hepatitis A   Alabama    1969      49              380        3412450
5  Hepatitis A   Alabama    1970      51              413        3444165
6  Hepatitis A   Alabama    1971      51              378        3481798
6 rows
the_disease = "Smallpox"
dat <- us_contagious_diseases %>% 
   filter(!state%in%c("Hawaii","Alaska") & disease == the_disease & !weeks_reporting<10) %>% 
   mutate(rate = count / population * 10000) %>% 
   mutate(state = reorder(state, rate))

dat %>% ggplot(aes(year, state, fill = rate)) + 
  geom_tile(color = "grey50") + 
  scale_x_continuous(expand=c(0,0)) + 
  scale_fill_gradientn(colors = brewer.pal(9, "Reds"), trans = "sqrt") + 
  theme_minimal() + 
  theme(panel.grid = element_blank()) + 
  ggtitle(the_disease) + 
  ylab("") + 
  xlab("")

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  1. Time series plot - measles and smallpox

The sample code given creates a time series plot showing the rate of measles cases per population by state. We are going to again modify this plot to look at smallpox cases instead.

library(dplyr)
library(ggplot2)
library(dslabs)
library(RColorBrewer)
data(us_contagious_diseases)

the_disease = "Measles"
dat <- us_contagious_diseases %>%
   filter(!state%in%c("Hawaii","Alaska") & disease == the_disease) %>%
   mutate(rate = count / population * 10000) %>%
   mutate(state = reorder(state, rate))
str(dat)
## 'data.frame':    3724 obs. of  7 variables:
##  $ disease        : Factor w/ 7 levels "Hepatitis A",..: 2 2 2 2 2 2 2 2 2 2 ...
##  $ state          : Factor w/ 51 levels "Mississippi",..: 9 9 9 9 9 9 9 9 9 9 ...
##   ..- attr(*, "scores")= num [1:51(1d)] 9.27 NA 24.15 9.37 19.16 ...
##   .. ..- attr(*, "dimnames")=List of 1
##   .. .. ..$ : chr  "Alabama" "Alaska" "Arizona" "Arkansas" ...
##  $ year           : num  1928 1929 1930 1931 1932 ...
##  $ weeks_reporting: int  52 49 52 49 41 51 52 49 40 49 ...
##  $ count          : num  8843 2959 4156 8934 270 ...
##  $ population     : num  2589923 2619131 2646248 2670818 2693027 ...
##  $ rate           : num  34.1 11.3 15.7 33.5 1 ...
avg <- us_contagious_diseases %>%
  filter(disease==the_disease) %>% group_by(year) %>%
  summarize(us_rate = sum(count, na.rm=TRUE)/sum(population, na.rm=TRUE)*10000)

dat %>% ggplot() +
  geom_line(aes(year, rate, group = state),  color = "grey50", 
            show.legend = FALSE, alpha = 0.2, size = 1) +
  geom_line(mapping = aes(year, us_rate),  data = avg, size = 1, color = "black") +
  scale_y_continuous(trans = "sqrt", breaks = c(5,25,125,300)) + 
  ggtitle("Cases per 10,000 by state") + 
  xlab("") + 
  ylab("") +
  geom_text(data = data.frame(x=1955, y=50), mapping = aes(x, y, label="US average"), color="black") + 
  geom_vline(xintercept=1963, col = "blue")

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library(dplyr)
library(ggplot2)
library(dslabs)
library(RColorBrewer)
data(us_contagious_diseases)

the_disease = "Smallpox"
dat <- us_contagious_diseases %>%
   filter(!state%in%c("Hawaii","Alaska") & disease == the_disease & !weeks_reporting<10) %>%
   mutate(rate = count / population * 10000) %>%
   mutate(state = reorder(state, rate))
str(dat)
## 'data.frame':    1014 obs. of  7 variables:
##  $ disease        : Factor w/ 7 levels "Hepatitis A",..: 7 7 7 7 7 7 7 7 7 7 ...
##  $ state          : Factor w/ 51 levels "Rhode Island",..: 17 17 17 17 17 17 17 17 17 17 ...
##   ..- attr(*, "scores")= num [1:51(1d)] 0.382 NA 2.011 0.805 0.924 ...
##   .. ..- attr(*, "dimnames")=List of 1
##   .. .. ..$ : chr  "Alabama" "Alaska" "Arizona" "Arkansas" ...
##  $ year           : num  1928 1929 1930 1931 1932 ...
##  $ weeks_reporting: int  51 52 52 52 52 52 52 52 51 52 ...
##  $ count          : num  341 378 192 295 467 82 23 42 12 54 ...
##  $ population     : num  2589923 2619131 2646248 2670818 2693027 ...
##  $ rate           : num  1.317 1.443 0.726 1.105 1.734 ...
avg <- us_contagious_diseases %>%
  filter(disease==the_disease) %>% group_by(year) %>%
  summarize(us_rate = sum(count, na.rm=TRUE)/sum(population, na.rm=TRUE)*10000)

dat %>% ggplot() +
  geom_line(aes(year, rate, group = state),  color = "grey50", 
            show.legend = FALSE, alpha = 0.2, size = 1) +
  geom_line(mapping = aes(year, us_rate),  data = avg, size = 1, color = "black") +
  scale_y_continuous(trans = "sqrt", breaks = c(5,25,125,300)) + 
  ggtitle("Cases per 10,000 by state") + 
  xlab("") + 
  ylab("") +
  geom_text(data = data.frame(x=1955, y=50), mapping = aes(x, y, label="US average"), color="black") + 
  geom_vline(xintercept=1963, col = "blue")

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  1. Time series plot - all diseases in California

Now we are going to look at the rates of all diseases in one state. Again, you will be modifying the sample code to produce the desired plot. - For the state of California, make a time series plot showing rates for all diseases. - Include only years with 10 or more weeks reporting. - Use a different color for each disease.

library(dplyr)
library(ggplot2)
library(dslabs)
library(RColorBrewer)
data(us_contagious_diseases)

us_contagious_diseases %>% filter(state=="California" & !weeks_reporting<10) %>% 
  group_by(year, disease) %>%
  summarize(rate = sum(count)/sum(population)*10000) %>%
  ggplot(aes(year, rate,color=disease)) + 
  geom_line()

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  1. Time series plot - all diseases in the United States

Now we are going to make a time series plot for the rates of all diseases in the United States. For this exercise, we have provided less sample code - you can take a look at the previous exercise to get you started. - Compute the US rate by using summarize to sum over states. - The US rate for each disease will be the total number of cases divided by the total population. - Remember to convert to cases per 10,000. - You will need to filter for !is.na(population) to get all the data. - Plot each disease in a different color.

library(dplyr)
library(ggplot2)
library(dslabs)
library(RColorBrewer)
data(us_contagious_diseases)

us_contagious_diseases %>% filter(!is.na(population)) %>% 
  group_by(year, disease) %>%
  summarize(rate=sum(count)/sum(population)*10000) %>%
  ggplot(aes(year, rate,color=disease)) + geom_line()

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Properties of Stars Exercises

Background
Astronomy is one of the oldest data-driven sciences. In the late 1800s, the director of the Harvard College Observatory hired women to analyze astronomical data, which at the time was done using photographic glass plates. These women became known as the Harvard Computers. They computed the position and luminosity of various astronomical objects such as stars and galaxies. (If you are interested, you can learn more about the Harvard Computers). Today, astronomy is even more of a data-driven science, with an inordinate amount of data being produced by modern instruments every day.

In the following exercises we will analyze some actual astronomical data to inspect properties of stars, their absolute magnitude (which relates to a star’s luminosity, or brightness), temperature and type (spectral class).

Libraries and Options

#update.packages()
library(tidyverse)
library(dslabs)
data(stars)
options(digits = 3)   # report 3 significant digits

Question 1
Load the stars data frame from dslabs. This contains the name, absolute magnitude, temperature in degrees Kelvin, and spectral class of selected stars. Absolute magnitude (shortened in these problems to simply “magnitude”) is a function of star luminosity, where negative values of magnitude have higher luminosity.

# What is the mean magnitude?
mean(stars$magnitude)
## [1] 4.26
# What is the standard deviation of magnitude?
sd(stars$magnitude)
## [1] 7.35

Question 2
Make a density plot of the magnitude.

stars %>%
  ggplot(aes(magnitude)) +
  geom_density()

# How many peaks are there in the data?
# A: 2

Question 3
Examine the distribution of star temperature. Which of these statements best characterizes the temperature distribution?

stars %>%
  ggplot(aes(temp)) +
  geom_density()

# How many peaks are there in the data?
# A: 2

Question 4
Make a scatter plot of the data with temperature on the x-axis and magnitude on the y-axis and examine the relationship between the variables. Recall that lower magnitude means a more luminous (brighter) star.

stars %>%
  ggplot(aes(x=temp, y=magnitude)) +
  geom_point()

Question 5
For various reasons, scientists do not always follow straight conventions when making plots, and astronomers usually transform values of star luminosity and temperature before plotting. Flip the y-axis so that lower values of magnitude are at the top of the axis (recall that more luminous stars have lower magnitude) using scale_y_reverse. Take the log base 10 of temperature and then also flip the x-axis.
Fill in the blanks in the statements below to describe the resulting plot:
The brighest, highest temperature stars are in the ______________ corner of the plot.

stars %>%
  ggplot(aes(x=log10(temp), y=magnitude)) +
  scale_y_reverse() +
  scale_x_reverse() +
  geom_point()

Question 6
The trends you see allow scientists to learn about the evolution and lifetime of stars. The primary group of stars to which most stars belong (see question 4) we will call the main sequence stars. Most stars belong to this main sequence, however some of the more rare stars are classified as old and evolved stars. These stars tend to be hotter stars, but also have low luminosity, and are known as white dwarfs.

How many white dwarfs are there in our sample?
A: 4

Question 7
Consider stars which are not part of the Main Group but are not old/evolved (white dwarf) stars. These stars must also be unique in certain ways and are known as giants. Use the plot from Question 5 to estimate the average temperature of a giant.

Which of these temperatures is closest to the average temperature of a giant?: A: 5000K

Question 8
We can now identify whether specific stars are main sequence stars, red giants or white dwarfs. Add text labels to the plot to answer these questions. You may wish to plot only a selection of the labels, repel the labels, or zoom in on the plot in RStudio so you can locate specific stars.
Fill in the blanks in the statements below:

library(ggrepel)
stars %>%
  ggplot(aes(x=log10(temp), y=magnitude, label=star)) +
  scale_y_reverse() +
  scale_x_reverse() +
  geom_point() +
  geom_text(aes(label=star)) +
  geom_text_repel()

# The least lumninous star in the sample with a surface temperature over 5000K is _________.
# A: van Maanens Star
# The two stars with lowest temperature and highest luminosity are known as supergiants. The two supergiants in this dataset are ____________.
# A: Betelgeuse and Antares
# The Sun is a ______________.
# A: main sequence star
stars %>% 
  filter(star=='Sun') %>%
  select_all()
##   star magnitude temp type
## 1  Sun       4.8 5840    G

Question 9
Remove the text labels and color the points by star type. This classification describes the properties of the star’s spectrum, the amount of light produced at various wavelengths.

stars %>%
  ggplot(aes(x=log10(temp), y=magnitude, color=type)) +
  scale_y_reverse() +
  scale_x_reverse() +
  geom_point()

# Which star type has the lowest temperature?

Climate Change Exercises

Background

The planet’s surface temperature is increasing due to human greenhouse gas emissions, and this global warming and carbon cycle disruption is wreaking havoc on natural systems. Living systems that depend on current temperature, weather, currents and carbon balance are jeopardized, and human society will be forced to contend with widespread economic, social, political and environmental damage as the temperature continues to rise. Although most countries recognize that global warming is a crisis and that humans must act to limit its effects, little action has been taken to limit or reverse human impact on the climate.

One limitation is the spread of misinformation related to climate change and its causes, especially the extent to which humans have contributed to global warming. In these exercises, we examine the relationship between global temperature changes, greenhouse gases and human carbon emissions using time series of actual atmospheric and ice core measurements from the National Oceanic and Atmospheric Administration (NOAA) and Carbon Dioxide Information Analysis Center (CDIAC).

Libraries and Options

#update.packages()
library(tidyverse)
library(dslabs)
data(temp_carbon)
data(greenhouse_gases)
data(historic_co2)

Question 1
Load the temp_carbon dataset from dslabs, which contains annual global temperature anomalies (difference from 20th century mean temperature in degrees Celsius), temperature anomalies over the land and ocean, and global carbon emissions (in metric tons). Note that the date ranges differ for temperature and carbon emissions.

Which of these code blocks return the latest year for which carbon emissions are reported?

str(temp_carbon)
## 'data.frame':    268 obs. of  5 variables:
##  $ year            : num  1880 1881 1882 1883 1884 ...
##  $ temp_anomaly    : num  -0.11 -0.08 -0.1 -0.18 -0.26 -0.25 -0.24 -0.28 -0.13 -0.09 ...
##  $ land_anomaly    : num  -0.48 -0.4 -0.48 -0.66 -0.69 -0.56 -0.51 -0.47 -0.41 -0.31 ...
##  $ ocean_anomaly   : num  -0.01 0.01 0 -0.04 -0.14 -0.17 -0.17 -0.23 -0.05 -0.02 ...
##  $ carbon_emissions: num  236 243 256 272 275 277 281 295 327 327 ...
temp_carbon %>%
    .$year %>%
    max()
## [1] 2018
temp_carbon %>%
    filter(!is.na(carbon_emissions)) %>%
    pull(year) %>%
    max()
## [1] 2014
#temp_carbon %>%
#    filter(!is.na(carbon_emissions)) %>%
#    max(year)
temp_carbon %>%
    filter(!is.na(carbon_emissions)) %>%
    .$year %>%
    max()
## [1] 2014
temp_carbon %>%
    filter(!is.na(carbon_emissions)) %>%
    select(year) %>%
    max()
## [1] 2014
#temp_carbon %>%
#    filter(!is.na(carbon_emissions)) %>%
#    max(.$year)

Question 2
Inspect the difference in carbon emissions in temp_carbon from the first available year to the last available year.

# What is the first year for which carbon emissions (carbon_emissions) data are available?
year_min <- temp_carbon %>%
  filter(!is.na(carbon_emissions)) %>%
  .$year %>%
  min()
# What is the last year for which carbon emissions data are available?
year_max <- temp_carbon %>%
  filter(!is.na(carbon_emissions)) %>%
  .$year %>%
  max()
# How many times larger were carbon emissions in the last year relative to the first year?
ratio <- temp_carbon %>%
  filter(year %in% c(year_min, year_max)) %>%
  .$carbon_emissions
#A:
ratio[1] / ratio[2]
## [1] 3285
# Scatter plot
temp_carbon %>%
  filter(!is.na(carbon_emissions)) %>%
  ggplot(aes(x=year, y=carbon_emissions)) +
  geom_point()

Question 3
Inspect the difference in temperature in temp_carbon from the first available year to the last available year.

# What is the first year for which global temperature anomaly (temp_anomaly) data are available?
year_min <- temp_carbon %>%
  filter(!is.na(temp_anomaly)) %>%
  .$year %>%
  min()
year_min
## [1] 1880
# What is the last year for which global temperature anomaly data are available?
year_max <- temp_carbon %>%
  filter(!is.na(temp_anomaly)) %>%
  .$year %>%
  max()
year_max
## [1] 2018
# How many degrees Celsius has temperature increased over the date range?
diff <- temp_carbon %>%
  filter(year %in% c(year_min, year_max)) %>%
  .$temp_anomaly
#A:
diff
## [1] -0.11  0.82
diff[1] - diff[2]
## [1] -0.93

Question 4 Create a time series line plot of the temperature anomaly. Only include years where temperatures are reported. Save this plot to the object p.
Which command adds a blue horizontal line indicating the 20th century mean temperature?

p <- temp_carbon %>%
  filter(!is.na(temp_anomaly)) %>%
  ggplot(aes(year, temp_anomaly)) +
  geom_line() + 
  geom_hline(aes(yintercept=0), color='blue')
p

Question 5
Continue working with p, the plot created in the previous question.

Change the y-axis label to be “Temperature anomaly (degrees C)”. Add a title, “Temperature anomaly relative to 20th century mean, 1880-2018”. Also add a text layer to the plot: the x-coordinate should be 2000, the y-coordinate should be 0.05, the text should be “20th century mean”, and the text color should be blue.

q <- temp_carbon %>%
  filter(!is.na(temp_anomaly)) %>%
  ggplot(aes(year, temp_anomaly)) +
  geom_line() + 
  geom_hline(aes(yintercept=0), color='blue') +
  ylab("Temperature anomaly (degrees C)") +
  ggtitle("Temperature anomaly relative to 20th century mean, 1880-2018") +
  geom_text(aes(x=2000, y=0.05, label="20th century mean"), col='blue')
q

Question 6

When was the earliest year with a temperature above the 20th century mean?

year_min <- temp_carbon %>%
  filter(!is.na(temp_anomaly) & temp_anomaly>0) %>%
  .$year %>%
  min()
year_min
## [1] 1939

When was the last year with an average temperature below the 20th century mean?

year_max <- temp_carbon %>%
  filter(!is.na(temp_anomaly) & temp_anomaly<0) %>%
  .$year %>%
  max()
year_max
## [1] 1976

In what year did the temperature anomaly exceed 0.5 degrees Celsius for the first time?

year_ <- temp_carbon %>%
  filter(!is.na(temp_anomaly) & temp_anomaly>0.5) %>%
  .$year %>%
  min()
year_
## [1] 1997

Question 7 Add layers to the previous plot to include line graphs of the temperature anomaly in the ocean (ocean_anomaly) and on land (land_anomaly). Assign different colors to the lines. Compare the global temperature anomaly to the land temperature anomaly and ocean temperature anomaly.

Which region has the largest 2018 temperature anomaly relative to the 20th century mean?

temp_carbon %>%
  filter(!is.na(temp_anomaly)) %>%
  ggplot(aes(year, temp_anomaly)) +
  geom_line(col='red') + 
  geom_hline(aes(yintercept=0), color='blue') +
  xlim(c(1880, 2018)) +
  ylab("Temperature anomaly (degrees C)") +
  ggtitle("Temperature anomaly relative to 20th century mean, 1880-2018") +
  geom_text(aes(x=2000, y=0.05, label="20th century mean"), col='blue') +
  geom_line(aes(year, ocean_anomaly), col='cyan') +
  geom_line(aes(year, land_anomaly), col='green')

Question 8 A major determinant of Earth’s temperature is the greenhouse effect. Many gases trap heat and reflect it towards the surface, preventing heat from escaping the atmosphere. The greenhouse effect is vital in keeping Earth at a warm enough temperature to sustain liquid water and life; however, changes in greenhouse gas levels can alter the temperature balance of the planet.

The greenhouse_gases data frame from dslabs contains concentrations of the three most significant greenhouse gases: carbon dioxide ( CO2 , abbreviated in the data as co2), methane ( CH4 , ch4 in the data), and nitrous oxide ( N2O , n2o in the data). Measurements are provided every 20 years for the past 2000 years.

str(greenhouse_gases)
## 'data.frame':    300 obs. of  3 variables:
##  $ year         : num  20 40 60 80 100 120 140 160 180 200 ...
##  $ gas          : chr  "CO2" "CO2" "CO2" "CO2" ...
##  $ concentration: num  278 278 277 277 278 ...

Complete the code outline below to make a line plot of concentration on the y-axis by year on the x-axis. Facet by gas, aligning the plots vertically so as to ease comparisons along the year axis. Add a vertical line with an x-intercept at the year 1850, noting the unofficial start of the industrial revolution and widespread fossil fuel consumption. Note that the units for ch4 and n2o are ppb while the units for co2 are ppm.

greenhouse_gases %>%
    ggplot(aes(year, concentration)) +
    geom_line() +
    facet_grid(gas ~ ., scales = "free") +
    geom_vline(xintercept = 1850, col='red') +
    ylab("Concentration (ch4/n2o ppb, co2 ppm)") +
    ggtitle("Atmospheric greenhouse gas concentration by year, 0-2000")

Question 10 Make a time series line plot of carbon emissions (carbon_emissions) from the temp_carbon dataset. The y-axis is metric tons of carbon emitted per year.

temp_carbon %>%
  filter(!is.na(carbon_emissions)) %>%
  ggplot(aes(year, carbon_emissions)) +
  geom_line()

Question 11
We saw how greenhouse gases have changed over the course of human history, but how has CO2 (co2 in the data) varied over a longer time scale? The historic_co2 data frame in dslabs contains direct measurements of atmospheric co2 from Mauna Loa since 1959 as well as indirect measurements of atmospheric co2 from ice cores dating back 800,000 years.

Make a line plot of co2 concentration over time (year), coloring by the measurement source (source). Save this plot as co2_time for later use.

co2_time <- historic_co2 %>%
  filter(!is.na(co2)) %>%
  ggplot(aes(year, co2, col=source)) +
  geom_line() +
  ggtitle("Atmospheric CO2 concentration, -800,000 BC to today") +
  ylab("co2 (ppmv)")
co2_time

Question 12
One way to differentiate natural co2 oscillations from today’s manmade co2 spike is by examining the rate of change of co2. The planet is affected not only by the absolute concentration of co2 but also by its rate of change. When the rate of change is slow, living and nonliving systems have time to adapt to new temperature and gas levels, but when the rate of change is fast, abrupt differences can overwhelm natural systems. How does the pace of natural co2 change differ from the current rate of change?

Use the co2_time plot saved above. Change the limits as directed to investigate the rate of change in co2 over various periods with spikes in co2 concentration.